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Sorted list of sums of 3 prices in minor currency units for a currency that has a 2-decimal minor unit, such that the riddle "sum of prices equals product of prices" has a solution, with prices expressed as floating point numbers with 2 decimals.
+0
0
525, 540, 546, 549, 555, 561, 567, 570, 585, 588, 600, 612, 630, 642, 660, 660, 663, 675, 726, 735, 744, 750, 759, 765, 783, 792, 798, 810, 819, 825, 840, 840, 891, 897, 900, 930, 945, 957, 966, 966, 975, 981, 996, 1050, 1050, 1071, 1080, 1092, 1125, 1134, 1155, 1155, 1170
OFFSET
1,1
COMMENTS
The sequence has 622 terms. See linked files for all solutions.
A natural number s occurs k times in the list if there exist k multisets {x,y,z} of natural numbers with s = x + y + z and 10000*s = x*y*z.
LINKS
EXAMPLE
a(1) = 525 because 1.50 + 1.75 + 2.00 = 1.50*1.75*2.00 = 5.25 is the solution with minimum sum;
a(15) = a(16) = 660 because there are 2 solutions:
0.80 + 2.50 + 3.30 = 0.80*2.50*3.30 = 6.60 and
1.10 + 1.50 + 4.00 = 1.10*1.50*4.00 = 6.60;
a(31) = a(32) = 840:
0.60 + 2.80 + 5.00 = 0.60*2.80*5.00 = 8.40 and
1.00 + 1.40 + 6.00 = 1.00*1.40*6.00 = 8.40;
a(622) = 100030002 is the largest term:
0.01 + 100.01 + 1000200.00 = 0.01*100.01*1000200.00 = 1000300.02.
CROSSREFS
KEYWORD
nonn,base,fini,full,new
AUTHOR
Hugo Pfoertner and Markus Sigg, Mar 02 2025
STATUS
approved
a(1) = 1; thereafter the sequence is extended by iteratively appending the run length transform of the reverse of the sequence thus far.
+0
0
1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 3, 1, 1, 1, 2, 1, 3, 1, 1, 1, 3, 1, 1, 1, 2, 1, 3, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 2, 1, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 2, 1, 3, 1, 3, 1, 3, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 3, 1, 3
OFFSET
1,3
COMMENTS
The run length transform replaces each run of consecutive equal values with a single value representing the length of that run.
Is 2 the greatest even number in the sequence?
EXAMPLE
Irregular triangle begins:
1;
1;
2;
1,2;
1,1,1,2;
1,3,1,1,1,2;
1,3,1,1,1,3,1,1,1,2;
CROSSREFS
Cf. A306211.
KEYWORD
tabf,nonn,new
AUTHOR
Neal Gersh Tolunsky, Feb 27 2025
STATUS
approved
a(0) = 4; for n > 0, a(n) = a(n-1) + n if G = 1 or a(n) = n/G if G > 1, where G = gcd(a(n-1), n).
+0
0
4, 5, 7, 10, 2, 7, 13, 20, 2, 11, 21, 32, 3, 16, 7, 22, 8, 25, 43, 62, 10, 31, 53, 76, 6, 31, 57, 9, 37, 66, 5, 36, 8, 41, 75, 7, 43, 80, 19, 58, 20, 61, 103, 146, 22, 67, 113, 160, 3, 52, 25, 76, 13, 66, 9, 64, 7, 64, 29, 88, 15, 76, 31, 94, 32, 97, 163, 230, 34, 103, 173, 244
OFFSET
0,1
COMMENTS
If a(n) < n for some n, then a(n+1) > n+1.
If a(n) > n, a(n+1) > n+1, and a(n+2) > n+2 for some n, then a(n+3) < n+3.
1 < a(n) for all n.
sqrt(n/6) < a(n) <= 7n/2 - 9/2 for all n.
a(p)>p for all primes p.
If one were to use the same rule to generate this sequence with any other initial value that is congruent to 4 or 8 (mod 12), that sequence would agree with this one for all n>3.
If one were to use the same rule to generate this sequence with an initial term that is not congruent to 4 or 8 (mod 12), then it would output the number 1 before the 5th term. When a sequence follows a(n)’s rules and outputs the number 1 at some index k, one gets the following quasi-periodic behavior: 1, k+2, 1, k+4, 1, k+6, etc., and are as such “boring” sequences.
EXAMPLE
a(12) = 3 and gcd(3, 13) = 1, so a(13) = 3 + 13 = 16. gcd(16, 14) = 2, so a(14) = 14/2 = 7.
PROG
(PARI) lista(nn) = my(v = vector(nn)); v[1] = 4; for (n=2, nn, my(g=gcd(v[n-1], n-1)); if (g==1, v[n] = v[n-1] + n-1, v[n] = (n-1)/g); ); v; \\ Michel Marcus, Feb 26 2025
CROSSREFS
Similar to A133058, A091508.
KEYWORD
nonn,new
AUTHOR
Sam Chapman, Feb 24 2025
STATUS
approved
Perimeter of the Sierpiński carpet at iteration n.
+0
0
4, 16, 80, 496, 3536, 26992, 212048, 1684720, 13442768, 107437168, 859182416, 6872514544, 54977282000, 439809752944, 3518452514384, 28147543587568, 225180119118032, 1801440264196720, 14411520047331152, 115292154179921392, 922337214843187664, 7378697662956950896, 59029581136289955920, 472236648588222693616
OFFSET
0,1
COMMENTS
Carpet n has an overall size 3^n X 3^n and the perimeter here includes the perimeter of all holes within it.
Carpet n=0 is a unit square and has perimeter a(0) = 4.
Carpet n can be constructed by arranging 8 copies of carpet n-1 in a square with a hole in the middle,
X X X
X X
X X X
There are no gaps in each side so 2 sides of each n-1 are now not on the perimeter so a(n) = 8*a(n-1) - 16*3^(n-1).
An equivalent construction is to replace each of the 8^(n-1) unit squares of carpet n-1 with a 3 X 3 block of unit squares with a hole in the middle, so that a(n) = 3*a(n-1) + 4*8^(n-1).
A fractal is obtained by scaling the whole carpet down to a unit square and its scaled perimeter a(n)/3^n -> oo shows the perimeter is infinite even though the area is bounded.
LINKS
Michael Small, Brendan Florio, and Phillip Donald Fawell, The use of the perimeter area method to calculate the fractal dimension of aggregates, see section 3.2 equation (27) where a(n) = P_s(n+1) with scale factor g_1 = 1.
Eric Weisstein's World of Mathematics, Sierpiński Carpet
FORMULA
a(n) = (4/5)*(4*3^n + 8^n).
a(n) = A365606(n+1) - 4.
EXAMPLE
For n=0, a(0) = 4, the geometric representation is a square.
For n=3, a(3) = 496.
PROG
(Python)
a = lambda n: (4 * (4 * 3**n + 8**n)) // 5
CROSSREFS
Cf. A113210 (fractal dimension).
KEYWORD
nonn,easy,new
AUTHOR
Jakub Buczak, Feb 26 2025
STATUS
approved
Last part of the section-sum partition of the prime indices of n.
+0
0
0, 1, 2, 1, 3, 3, 4, 1, 2, 4, 5, 1, 6, 5, 5, 1, 7, 2, 8, 1, 6, 6, 9, 1, 3, 7, 2, 1, 10, 6, 11, 1, 7, 8, 7, 3, 12, 9, 8, 1, 13, 7, 14, 1, 2, 10, 15, 1, 4, 3, 9, 1, 16, 2, 8, 1, 10, 11, 17, 1, 18, 12, 2, 1, 9, 8, 19, 1, 11, 8, 20, 1, 21, 13, 3, 1, 9, 9, 22, 1, 2
OFFSET
1,3
COMMENTS
A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.
The section-sum partition (A381436) of a multiset or partition y is defined as follows: (1) determine and remember the sum of all distinct parts, (2) remove one instance of each distinct part, (3) repeat until no parts are left. The remembered values comprise the section-sum partition. For example, starting with (3,2,2,1,1) we get (6,3).
Equivalently, the k-th part of the section-sum partition is the sum of all (distinct) parts that appear at least k times. Compare to the definition of the conjugate of a partition, where we count parts >= k.
The conjugate of a section-sum partition is a Look-and-Say partition; see A048767, union A351294, count A239455.
FORMULA
a(n) = A055396(A381431(n)).
EXAMPLE
The prime indices of 972 are {1,1,2,2,2,2,2}, with section-sum partition (3,3,2,2,2), so a(972) = 2.
MATHEMATICA
prix[n_]:=If[n==1, {}, Flatten[Cases[FactorInteger[n], {p_, k_}:>Table[PrimePi[p], {k}]]]];
egs[y_]:=If[y=={}, {}, Table[Total[Select[Union[y], Count[y, #]>=i&]], {i, Max@@Length/@Split[y]}]];
Table[If[n==1, 0, Last[egs[prix[n]]]], {n, 100}]
CROSSREFS
Positions of first appearances are A008578.
The length of this partition is A051903.
The conjugate version is A051904.
For first instead of last part we get A066328.
These partitions are counted by A239455, complement A351293.
Positions of 1 are A360013, complement A381439.
This is the least prime index of A381431 (see A381432, A381433, A381434, A381435).
This is the last part of row n of A381436 (see A381440, A048767, A351294, A351295).
Counting partitions by this statistic gives A381438.
A000040 lists the primes, differences A001223.
A055396 gives least prime index, greatest A061395.
A056239 adds up prime indices, row sums of A112798.
A122111 represents conjugation in terms of Heinz numbers.
Set multipartitions: A050320, A089259, A116540, A270995, A296119, A318360, A318361.
KEYWORD
nonn,new
AUTHOR
Gus Wiseman, Feb 28 2025
STATUS
approved
Triangle read by rows where T(n>0,k>0) is the number of integer partitions of n whose section-sum partition ends with k.
+0
0
1, 1, 1, 1, 0, 2, 2, 1, 0, 2, 3, 1, 0, 0, 3, 4, 1, 2, 0, 0, 4, 7, 2, 1, 0, 0, 0, 5, 9, 4, 1, 2, 0, 0, 0, 6, 13, 4, 4, 1, 0, 0, 0, 0, 8, 18, 6, 3, 2, 3, 0, 0, 0, 0, 10, 26, 9, 5, 2, 2, 0, 0, 0, 0, 0, 12, 32, 12, 8, 4, 2, 4, 0, 0, 0, 0, 0, 15
OFFSET
1,6
COMMENTS
The section-sum partition (A381436) of a multiset or partition y is defined as follows: (1) determine and remember the sum of all distinct parts, (2) remove one instance of each distinct part, (3) repeat until no parts are left. The remembered values comprise the section-sum partition. For example, starting with (3,2,2,1,1) we get (6,3).
Equivalently, the k-th part of the section-sum partition is the sum of all (distinct) parts that appear at least k times. Compare to the definition of the conjugate of a partition, where we count parts >= k.
The conjugate of a section-sum partition is a Look-and-Say partition; see A048767, union A351294, count A239455.
EXAMPLE
Triangle begins:
1
1 1
1 0 2
2 1 0 2
3 1 0 0 3
4 1 2 0 0 4
7 2 1 0 0 0 5
9 4 1 2 0 0 0 6
13 4 4 1 0 0 0 0 8
18 6 3 2 3 0 0 0 0 10
26 9 5 2 2 0 0 0 0 0 12
32 12 8 4 2 4 0 0 0 0 0 15
47 16 11 4 3 2 0 0 0 0 0 0 18
60 23 12 8 3 2 5 0 0 0 0 0 0 22
79 27 20 7 9 4 3 0 0 0 0 0 0 0 27
Row n = 9 counts the following partitions:
(711) (522) (333) (441) . . . . (9)
(6111) (4221) (3321) (81)
(5211) (3222) (32211) (72)
(51111) (22221) (222111) (63)
(4311) (621)
(42111) (54)
(411111) (531)
(33111) (432)
(321111)
(3111111)
(2211111)
(21111111)
(111111111)
MATHEMATICA
egs[y_]:=If[y=={}, {}, Table[Total[Select[Union[y], Count[y, #]>=i&]], {i, Max@@Length/@Split[y]}]];
Table[Length[Select[IntegerPartitions[n], k==Last[egs[#]]&]], {n, 15}, {k, n}]
CROSSREFS
Last column (k=n) is A000009.
Row sums are A000041.
Row sums without the last column (k=n) are A047967.
For first instead of last part we have A116861, rank A066328.
First column (k=1) is A241131 shifted right and starting with 1 instead of 0.
Using Heinz numbers, this statistic is given by A381437.
A122111 represents conjugation in terms of Heinz numbers.
A239455 counts section-sum partitions, complement A351293.
Set multipartitions: A050320, A089259, A116540, A270995, A296119, A318360, A318361.
Section-sum partition: A381431, A381432, A381433, A381434, A381435, A381436.
Look-and-Say partition: A048767, A351294, A351295, A381440.
KEYWORD
nonn,tabl,new
AUTHOR
Gus Wiseman, Mar 01 2025
STATUS
approved
a(n) is the least positive integer k for which k^2 + (k + n)^2 is a square.
+0
0
3, 6, 9, 12, 15, 18, 5, 24, 27, 30, 33, 36, 39, 10, 45, 48, 7, 54, 57, 60, 15, 66, 12, 72, 75, 78, 81, 20, 87, 90, 9, 96, 99, 14, 25, 108, 111, 114, 117, 120, 36, 30, 129, 132, 135, 24, 16, 144, 11, 150, 21, 156, 159, 162, 165, 40, 171, 174, 177, 180, 183, 18, 45
OFFSET
1,1
COMMENTS
a(n) is also the smallest short leg of a Pythagorean triangle where the difference between the two legs is n.
A289398(n) is the least integer m > n for which (n^2 + m^2)/2 is a square. This is equivalent to the least positive integer k for which (n^2 + (n + 2*k)^2)/2 = k^2 + (n + k)^2 is a square. From m = n + 2*k follows a(n) = (A289398(n) - n)/2.
LINKS
Eric Weisstein's World of Mathematics, Pythagorean Triple
FORMULA
a(n) = (A289398(n) - n)/2.
EXAMPLE
a(1) = 3 because 3^2 + (3 + 1)^2 = 5^2 and there is no smaller positive integer k than 3 with that property.
a(28) = 20 because 20^2 + (20 + 28)^2 = 52^2 and there is no smaller positive integer k than 20 with that property.
MAPLE
A379596:=proc(n)
local k;
for k do
if issqr(k^2+(k+n)^2) then
return k
fi
od
end proc;
seq(A379596(n), n=1..63);
PROG
(PARI) a(n) = my(k=1); while (!issquare(k^2 + (k + n)^2), k++); k; \\ Michel Marcus, Feb 15 2025
KEYWORD
nonn,easy,new
AUTHOR
Felix Huber, Feb 15 2025
STATUS
approved
Lexicographically earliest sequence of positive integers such that for any t and k, with k>=1, where t = a(n) = a(n+k) = a(n+2*k), only one occurrence of k, for a given t, appears anywhere in the sequence.
+0
0
1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 2, 1, 1, 2, 1, 1, 3, 2, 3, 2, 1, 3, 1, 2, 3, 2, 2, 1, 3, 3, 2, 1, 2, 3, 3, 3, 1, 1, 4, 1, 2, 3, 4, 3, 1, 3, 1, 4, 4, 2, 3, 2, 2, 4, 3, 4, 2, 4, 4, 2, 1, 4, 1, 3, 2, 2, 4, 5, 3, 1, 3, 3, 1, 4, 4, 2, 4, 4, 3, 1, 1, 2, 3, 3, 2, 5, 5, 3, 5, 2, 1, 3, 4, 5, 4, 1, 5, 4, 3, 1, 2, 4, 1, 4, 1, 5, 2, 2, 3, 3, 5, 5, 5, 4, 5, 1, 4, 3, 2, 5
OFFSET
1,4
COMMENTS
See A381599 for the index where n first appears, and A381598 for the index where three consecutive n's appears.
LINKS
EXAMPLE
a(1) = a(2) = a(3) = 1, which is the first appearance of three 1's separated by one term.
a(4) = 2 as 1 cannot be chosen as that would form a(2) = a(3) = a(4) = 1, but three 1's separated by one term has already appeared.
a(5) = 1, which also forms three 1's separated by two terms, a(1) = a(3) = a(5) = 1.
a(17) = 3 as 1 cannot be chosen as that would form a(15) = a(16) = a(17) = 1, but three 1's separated by one term has already appeared, while choosing 2 would form a(11) = a(14) = a(17) = 2, but three 2's separated by three terms has already appeared at a(4) = a(7) = a(10) = 2.
CROSSREFS
Cf. A381598 (triplets), A381599 (where n first appears), A370708 (indices where 1's appear), A281511, A229037.
KEYWORD
nonn,new
AUTHOR
Scott R. Shannon, Mar 01 2025
STATUS
approved
Index of first term of three consecutive n's in A381597.
+0
0
1, 9, 34, 147, 111, 359, 437, 389, 594, 826, 1102, 83317, 1789, 5142, 2931, 12671
OFFSET
1,2
COMMENTS
The terms vary greatly in size - after 5.2 million terms of A381597 no three consecutive 17's or 18's have appeared, although three consecutive 19's appear at index 6474. The largest known term is a(192) = 5135798.
CROSSREFS
KEYWORD
nonn,more,new
AUTHOR
Scott R. Shannon, Mar 01 2025
STATUS
approved
Index where n first appears in A381597.
+0
0
1, 4, 17, 39, 68, 124, 191, 286, 441, 577, 776, 1043, 1192, 1556, 1736, 2214, 2744, 3221, 3519, 4248, 5028, 5542, 6574, 7013, 8093, 8945, 10110, 11043, 12413, 13223, 14476, 15923, 17430, 18617, 20027, 21991, 24016, 25364, 27414, 29356, 31392, 32614, 35743, 37888, 40301, 42620, 45696, 47776, 51109, 53264, 56429, 58471, 61676, 64468, 69437, 72011, 75626
OFFSET
1,2
CROSSREFS
KEYWORD
nonn,new
AUTHOR
Scott R. Shannon, Mar 01 2025
STATUS
approved

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