Search: keyword:new
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0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
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OFFSET
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0,101
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COMMENTS
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More than the usual number of terms are shown in order to distinguish this sequence from A180160, from which it first differs at n = 100.
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LINKS
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MATHEMATICA
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A374097[n_] := #*(IntegerLength[n] - #) & [Total[Mod[IntegerDigits[n], 2]]];
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CROSSREFS
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KEYWORD
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nonn,base,easy,new
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AUTHOR
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STATUS
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approved
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A373917
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Triangle read by rows: T(n,k) = k*10 mod n, with n >= 1, k >= 0.
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+0
0
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0, 0, 0, 0, 1, 2, 0, 2, 0, 2, 0, 0, 0, 0, 0, 0, 4, 2, 0, 4, 2, 0, 3, 6, 2, 5, 1, 4, 0, 2, 4, 6, 0, 2, 4, 6, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 10, 8, 6, 4, 2, 0, 10, 8, 6, 4, 2, 0, 10, 7, 4, 1, 11, 8, 5, 2, 12, 9, 6, 3
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OFFSET
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1,6
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COMMENTS
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Each row n encodes a "division graph" used to determine m mod n (where m is an arbitrary nonnegative integer), using the method described in the Numberphile link (see also example).
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LINKS
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James Grime and Brady Haran, Solving Seven, Numberphile YouTube video, 2024.
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EXAMPLE
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Triangle begins:
n\k| 0 1 2 3 4 5 6 7 8 9
---------------------------------
1 | 0;
2 | 0, 0;
3 | 0, 1, 2;
4 | 0, 2, 0, 2;
5 | 0, 0, 0, 0, 0;
6 | 0, 4, 2, 0, 4, 2;
7 | 0, 3, 6, 2, 5, 1, 4;
8 | 0, 2, 4, 6, 0, 2, 4, 6;
9 | 0, 1, 2, 3, 4, 5, 6, 7, 8;
10 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
...
Suppose m = 3714289 and you want to determine m mod 7 (the example shown in the video).
Start with the first digit of m (3) and calculate T(7,3 mod 7) = T(7,3) = 2.
Add it to the next digit of m (7) and calculate T(7,(2+7) mod 7) = T(7,2) = 6.
Add it to the next digit of m (1) and calculate T(7,(6+1) mod 7) = T(7,0) = 0.
Add it to the next digit of m (4) and calculate T(7,(0+4) mod 7) = T(7,4) = 5.
Add it to the next digit of m (2) and calculate T(7,(5+2) mod 7) = T(7,0) = 0.
Add it to the next digit of m (8) and calculate T(7,(0+8) mod 7) = T(7,1) = 3.
Add it to the final digit of m (9) and calculate (3+9) mod 7 = 5, which corresponds to 3714289 mod 7.
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MATHEMATICA
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Table[Mod[Range[0, 10*(n-1), 10], n], {n, 15}]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A373889
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Square array read by ascending antidiagonals: T(k,n) is the cardinality of {(E is a proper finite subset of the natural numbers) such that E = {} or w_k(E) < min(E) <= max(E) <= n}, where w_k(E) = Sum_{i in E, i <> k} 1, with n, k >= 1.
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+0
0
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2, 1, 3, 1, 2, 4, 1, 2, 4, 6, 1, 2, 4, 7, 9, 1, 2, 3, 6, 11, 14, 1, 2, 3, 6, 10, 17, 22, 1, 2, 3, 5, 10, 17, 26, 35, 1, 2, 3, 5, 10, 16, 28, 40, 56, 1, 2, 3, 5, 8, 16, 26, 45, 62, 90, 1, 2, 3, 5, 8, 16, 26, 43, 71, 97, 145, 1, 2, 3, 5, 8, 13, 26, 42, 71, 111, 153, 234
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OFFSET
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1,1
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LINKS
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FORMULA
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T(k,n) = A000045(n-k+2) + 1, for k = 1 and n >= k;
T(k,n) = 2*(Sum_{i=0..k-2} binomial(n-k,i)*A000045(k-i)) + 2*binomial(n-k,k-1) + Sum_{j=1..n-k} binomial(j,n-j), for k >= 2 and n >= k;
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EXAMPLE
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The array begins:
k\n| 1 2 3 4 5 6 7 8 9 10 ...
----------------------------------------------
1 | 2, 3, 4, 6, 9, 14, 22, 35, 56, 90, ... = A001611 (from n = 2).
2 | 1, 2, 4, 7, 11, 17, 26, 40, 62, 97, ...
3 | 1, 2, 4, 6, 10, 17, 28, 45, 71, 111, ...
4 | 1, 2, 3, 6, 10, 16, 26, 43, 71, 116, ...
5 | 1, 2, 3, 5, 10, 16, 26, 42, 68, 111, ...
6 | 1, 2, 3, 5, 8, 16, 26, 42, 68, 110, ...
7 | 1, 2, 3, 5, 8, 13, 26, 42, 68, 110, ...
8 | 1, 2, 3, 5, 8, 13, 21, 42, 68, 110, ...
9 | 1, 2, 3, 5, 8, 13, 21, 34, 68, 110, ...
10 | 1, 2, 3, 5, 8, 13, 21, 34, 55, 110, ...
...
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MATHEMATICA
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A373889[k_, n_] := Which[n < k, Fibonacci[n+1], k == 1, Fibonacci[n-k+2] + 1, True, 2*Sum[Binomial[n-k, i]*Fibonacci[k-i], {i, 0, k-2}] + 2*Binomial[n-k, k-1] + Sum[Binomial[j, n-j], {j, n-k}]];
Table[A373889[k-n+1, n], {k, 15}, {n, k}]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A373196
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Maximal coefficient (in absolute value) in the numerator of C({1..n},x).
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+0
1
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1, 1, 2, 17, 444, 66559954, 14648786369948422, 791540878703169050660325841979096789557779, 1918013047695258943191946313451491492494186620117241479813740479213857275772347178176158
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OFFSET
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0,3
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LINKS
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FORMULA
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C({s},x) = Sum_{i in {s}} (C({s}-{i},x)*x^i)/(1 - Sum_{i in {s}} (x^i)) with C({},x) = 1.
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EXAMPLE
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C_x({1,2,3},x) = (-x^15 - 5*x^14 - 12*x^13 - 17*x^12 - 11*x^11 + 4*x^10 + 16*x^9 + 10*x^8 - 6*x^6)/(x^15 + 4*x^14 + 7*x^13 + 4*x^12 - 8*x^11 - 18*x^10 - 13*x^9 + 7*x^8 + 19*x^7 + 11*x^6 - 6*x^5 - 10*x^4 - 2*x^3 + 3*x^2 + 2*x - 1) with maximal coefficient abs(-17) in the numerator, so a(3) = 17.
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PROG
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(PARI)
C_x(s)={my(g=if(#s <1, 1, sum(i=1, #s, C_x(s[^i])*x^(s[i]))/(1-sum(i=1, #s, x^(s[i]))))); return(g)}
a(n)={vecmax(abs(Vec(numerator(C_x([1..n])))))}
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CROSSREFS
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KEYWORD
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nonn,new
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AUTHOR
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STATUS
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approved
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4, 6, 6, 1, 5, 2, 4, 4, 13, 9, 7, 2, 4, 6, 2, 7, 6, 4, 6, 1, 3, 7, 2, 10, 10, 5, 2, 4, 11, 6, 10, 2, 4, 6, 5, 6, 6, 4, 7, 5, 3, 8, 9, 0, 8, 4, 3, 4, 6, 8, 5, 3, 2, 8, 6, 1, 6, 6, 6, 9, 2, 17, 4, 13, 5, 7, 2, 5, 9, 4, 5, 1, 6, 3, 4, 2, 9, 10, 1, 3, 4, 4, 2, 12, 4, 5, 5, 6, 7, 12, 6, 6, 3, 4, 8, 3, 4, 9, 5, 7
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OFFSET
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38,1
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COMMENTS
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The offset is 38, because up to that point the terms are not integers.
It appears that in A373390 the primes up to and including 157 (which is A373390(335)) appear irregularly. The next prime is 163 = A373390(350). So there is a possibility that ignoring the first 347 or so terms of A373390 may make it easier to analyze. A formula or other explanation for the present sequence would be of great help.
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LINKS
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EXAMPLE
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The first 38 terms of the first differences of A373792, halved, are
0, 5, -5, 29/2, -15, 23/2, -23/2, 27, -19/2, -35/2, 41, -41/2, 3, 3, 61/2, -49/2, -65/2, 71/2, 67/2, 4, 4, 9, 3, -77/2, 97/2, 3, 7, -2, 2, 16, 3, 8, 1, 12, 0, -137/2, 161/2, 4,
and that final 4 is the leading term of the present sequence.
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KEYWORD
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nonn,new
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AUTHOR
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STATUS
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approved
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A374120
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Numbers k such that A113177(k) and A276085(k) are both multiples of 3, where A113177 and A276085 are fully additive with a(p) = Fibonacci(p) and a(p) = p#/p, respectively.
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+0
0
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1, 6, 8, 27, 35, 36, 48, 64, 77, 85, 91, 115, 125, 133, 155, 162, 187, 203, 205, 210, 216, 221, 235, 253, 259, 275, 280, 288, 299, 301, 323, 325, 341, 343, 355, 365, 371, 384, 395, 403, 413, 427, 437, 445, 451, 462, 469, 475, 485, 493, 510, 512, 515, 517, 533, 546, 565, 581, 589, 605, 611, 616, 629, 635, 667, 680
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OFFSET
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1,2
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COMMENTS
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A multiplicative semigroup: if m and n are in the sequence, then so is m*n.
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LINKS
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PROG
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CROSSREFS
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Indices of multiples of 3 in A374112.
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KEYWORD
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nonn,new
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AUTHOR
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STATUS
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approved
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A374119
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a(n) = 1 if A113177(n) and A276085(n) are both multiples of 3, otherwise 0, where A113177 and A276085 are fully additive with a(p) = Fibonacci(p) and a(p) = p#/p, respectively.
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+0
0
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1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1
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OFFSET
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1
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LINKS
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FORMULA
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PROG
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(PARI)
A113177(n) = if(n<=1, 0, my(f=factor(n)); sum(i=1, #f~, f[i, 2]*fibonacci(f[i, 1])));
A276085(n) = { my(f=factor(n)); sum(k=1, #f~, f[k, 2]*prod(i=1, primepi(f[k, 1]-1), prime(i))); };
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CROSSREFS
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Characteristic function of A374120.
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KEYWORD
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nonn,new
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AUTHOR
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STATUS
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approved
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1, 3, 4, 9, 12, 16, 25, 27, 35, 36, 48, 49, 55, 64, 65, 75, 77, 81, 85, 91, 95, 100, 105, 108, 115, 119, 121, 133, 140, 143, 144, 145, 147, 155, 161, 165, 169, 185, 187, 192, 195, 196, 203, 205, 209, 215, 217, 220, 221, 225, 231, 235, 243, 247, 253, 255, 256, 259, 260, 265, 273, 285, 287, 289, 295, 299, 300, 301, 305
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OFFSET
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1,2
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COMMENTS
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Numbers whose 2-adic valuation (A007814) is even, and the number of the prime factors (with multiplicity, A001222) and the 3-adic valuation (A007949) have the same parity.
A multiplicative semigroup: if m and n are in the sequence, then so is m*n.
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LINKS
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PROG
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CROSSREFS
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KEYWORD
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nonn,new
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AUTHOR
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STATUS
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approved
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A374115
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Numbers k such that A113177(k) and A276085(k) are not both even, where A113177 and A276085 are fully additive with a(p) = Fibonacci(p) and a(p) = p#/p, respectively.
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+0
0
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2, 5, 6, 7, 8, 10, 11, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 28, 29, 30, 31, 32, 33, 34, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 53, 54, 56, 57, 58, 59, 60, 61, 62, 63, 66, 67, 68, 69, 70, 71, 72, 73, 74, 76, 78, 79, 80, 82, 83, 84, 86, 87, 88, 89, 90, 92, 93, 94, 96, 97, 98, 99, 101, 102
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OFFSET
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1,1
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LINKS
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PROG
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CROSSREFS
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KEYWORD
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nonn,new
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AUTHOR
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STATUS
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approved
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A374113
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a(n) = 1 if A113177(n) and A276085(n) are both even, otherwise 0, where A113177 and A276085 are fully additive with a(p) = Fibonacci(p) and a(p) = p#/p, respectively.
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+0
0
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1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1
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OFFSET
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1
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COMMENTS
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a(n) = 1 if the 2-adic valuation of n is even, and the number of its prime factors (with multiplicity, A001222) and its 3-adic valuation (A007949) have the same parity, otherwise 0.
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LINKS
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FORMULA
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PROG
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(PARI)
A113177(n) = if(n<=1, 0, my(f=factor(n)); sum(i=1, #f~, f[i, 2]*fibonacci(f[i, 1])));
A276085(n) = { my(f=factor(n)); sum(k=1, #f~, f[k, 2]*prod(i=1, primepi(f[k, 1]-1), prime(i))); };
(PARI) A374113(n) = (!(valuation(n, 2)%2) && !((bigomega(n)-valuation(n, 3))%2));
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CROSSREFS
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Characteristic function of A374114, whose complement A374115 gives the indices of 0's.
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KEYWORD
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nonn,new
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AUTHOR
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STATUS
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approved
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