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A100982
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Number of admissible sequences of order j; related to 3x+1 problem and Wagon's constant.
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15
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1, 1, 2, 3, 7, 12, 30, 85, 173, 476, 961, 2652, 8045, 17637, 51033, 108950, 312455, 663535, 1900470, 5936673, 13472296, 39993895, 87986917, 257978502, 820236724, 1899474678, 5723030586, 12809477536, 38036848410, 84141805077, 248369601964
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OFFSET
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1,3
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COMMENTS
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Eric Roosendaal counted all admissible sequences up to order j=1000 (2005). Note: there is a typo in both Wagon and Chamberland in the definition of Wagon's constant 9.477955... The expression floor(1 + 2*i + i*log_2(3)) should be replaced by floor(1 + i + i*log_2(3)).
The length of all admissible sequences of order j is A020914(j). - T. D. Noe, Sep 11 2006
Conjecture: a(n) is given for each n > 3 by a formula using a(2)..a(n-1). This allows us to create an iterative algorithm which generates a(n) for each n > 6. This has been proved for each n <= 53. For higher values of n the algorithm must be slightly modified. - Mike Winkler, Jan 03 2018
Theorem 1: a(k) is given for each k > 1 by a formula using a(1)..a(k-1). Namely, a(1)=1 and a(k+1) = Sum_{m=1..k} (-1)^(m-1)*binomial(floor((k-m+1)*(log(3)/log(2))) + m - 1, m)*a(k-m+1)) for k >= 1. - Vladimir M. Zarubin, Sep 25 2015
Theorem 2: a(n) can be generated for each n > 2 algorithmically in a Pascal's triangle-like manner from the two starting values 0 and 1. This result is based on the fact that the Collatz residues (mod 2^k) can be evolved according to a binary tree. There is a direct connection with A076227, A056576 and A022921. - Mike Winkler, Sep 12 2017
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LINKS
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FORMULA
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A sequence s(k), where k=1, 2, ..., n, is *admissible* if it satisfies s(k)=3/2 exactly j times, s(k)=1/2 exactly n-j times, s(1)*s(2)*...*s(n) < 1 but s(1)*s(2)*...*s(m) > 1 for all 1 < m < n.
a(n) = (m+n-2)!/(m!*(n-2)!) - Sum_{i=2..n-1} binomial(floor((3*(n-i)+b)/2), n-i)*a(i), where m = floor((n-1)*log_2(3))-(n-1) and b assumes different integer values within the sum at intervals of 5 or 6 terms. (Conjecture)
a(n) = Sum_{k=n-1..A056576(n-1)} (k,n). (Theorem 2, cf. example)
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EXAMPLE
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The unique admissible sequence of order 1 is 3/2, 1/2.
The unique admissible sequence of order 2 is 3/2, 3/2, 1/2, 1/2.
The two admissible sequences of order 3 are 3/2, 3/2, 3/2, 1/2, 1/2 and 3/2, 3/2, 1/2, 3/2, 1/2.
a(13) = 8045 = binomial(floor(5*(13-2)/3), 13-2)
- Sum_{i=2..6} binomial(floor((3*(13-i)+0)/2), 13-i)*a(i)
- Sum_{i=7..11} binomial(floor((3*(13-i)-1)/2), 13-i)*a(i)
- Sum_{i=12..12} binomial(floor((3*(13-i)-2)/2), 13-i)*a(i)
= 31824 - 4368*1 - 3003*2 - 715*3 - 495*7 - 120*12 - 28*30 - 21*85 - 5*173 - 4*476 - 1*961 - 0*2652. (Conjecture)
The next table shows how Theorem 2 works. No entry is equal to zero.
n = 3 4 5 6 7 8 9 10 11 12 .. |A076227(k)=
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k = 2 | 1 | 1
k = 3 | 1 1 | 2
k = 4 | 2 1 | 3
k = 5 | 3 1 | 4
k = 6 | 3 4 1 | 8
k = 7 | 7 5 1 | 13
k = 8 | 12 6 1 | 19
k = 9 | 12 18 7 1 | 38
k = 10 | 30 25 8 1 | 64
k = 11 | 30 55 33 9 1 | 128
: | : : : : .. | :
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a(n) = 2 3 7 12 30 85 173 476 961 2652 .. |
The entries (k,n) in this table are generated by the rule (k+1,n) = (k,n) + (k,n-1). The last value of (k+1,n) is given by k+1 = A056576(n-1), or the highest value in column n is given twice only if A022921(n-2) = 2. Then a(n) is equal to the sum of the entries in column n. For n = 7 there is 1 = 0 + 1, 5 = 1 + 4, 12 = 5 + 7, 12 = 12 + 0. Therefore a(7) = 1 + 5 + 12 + 12 = 30. The sum of row k is equal to A076227(k). (End)
A tree view.
n-tree--A098294--ids-----paths-----------------a(n)
0 ._ 0 0 0 -
1 |_ 1 1 10 1
2 |_._ 2 2 1100 1
3 |_|_ 2 3-4 11010 - 11100 2
4 |_|_._ 3 5-7 1101100 - 1111000 3
5 |_|_|_ 3 8-14 11011010 - 11111000 7
6 |_|_|_._ 4 15-26 1101101100-1111110000 12
7 |_|_|_|_._ 5 27-56 ... 30
8 |_|_|_|_|_ 5 57-141 ... 85
...
For n>=1, the endpoints are at A098294(n) to the right.
(End)
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MATHEMATICA
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(* based on Eric Roosendaal's algorithm *) nn=100; Clear[x, y]; Do[x[i]=0, {i, 0, nn+1}]; x[1]=1; t=Table[Do[y[cnt]=x[cnt]+x[cnt-1], {cnt, p+1}]; Do[x[cnt]=y[cnt], {cnt, p+1}]; admis=0; Do[If[(p+1-cnt)*Log[3]<p*Log[2], admis=admis+x[cnt]; x[cnt]=0], {cnt, p+1}]; admis, {p, 2, nn}]; DeleteCases[t, 0] (* T. D. Noe, Sep 11 2006 *)
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PROG
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(PARI) /* translation of the above code from T. D. Noe */
{limit=100; n=1; x=y=vector(limit+1); x[1]=1; for(b=2, limit, for(c=2, b+1, y[c]=x[c]+x[c-1]); for(c=2, b+1, x[c]=y[c]); a_n=0; for(c=1, b+1, if((b+1-c)*log(3)<b*log(2), a_n=a_n+x[c]; x[c]=0)); if(a_n!=0, print(n" "a_n); n++))} \\ Mike Winkler, Feb 28 2015
(PARI) /* algorithm for the Conjecture */
{limit=53; zn=vector(limit); zn[2]=1; zn[3]=2; zn[4]=3; zn[5]=7; zn[6]=12; f=1; e1=-1; e2=-2; for(n=7, limit, m=floor((n-1)*log(3)/log(2))-(n-1); j=(m+n-2)!/(m!*(n-2)!); if(n>6*f, if(frac(n/2)==0, e=e1, e=e2)); if(frac((n-6 )/12)==0, f++; e1=e1+2); if(frac((n-12)/12)==0, f++; e2=e2+2); Sum=a=b=0; c=1; d=5; until(c>=n-1, for(i=2+a*5+b, 1+d+a*5, if(i>11 && frac((i+2)/6)==0, b++); delta=e-a; Sum=Sum+binomial(floor((3*(n-i)+delta)/2), n-i)*zn[i]; c++); a++; for(k=3, 50, if(n>=k*6 && a==k-1, d=k+3))); zn[n]=j-Sum; print(n" "zn[n]))} \\ Mike Winkler, Jan 03 2018
(PARI) /* cf. code for Theorem 2 */
{limit=100; /*or limit>100*/ p=q=vector(limit); c=2; w=log(3)/log(2); for(n=3, limit, p[1]=Sum=1; for(i=2, c, p[i]=p[i-1]+q[i]; Sum=Sum+p[i]); a_n=Sum; print(n" "a_n); for(i=1, c, q[i]=p[i]); d=floor(n*w)-floor((n-1)*w); if(d==2, c++)); } \\ Mike Winkler, Apr 14 2015
(PARI) /* algorithm for Theorem 1 */
n=20; a=vector(n); log32=log(3)/log(2);
{a[1]=1; for ( k=1, n-1, a[k+1]=sum( m=1, k, (-1)^(m-1)*binomial( floor( (k-m+1)*log32)+m-1, m)*a[k-m+1] ); print(k" "a[k]) );
(PARI) /* algorithm for Theorem 2 */
{limit=30; /*or limit>30*/ R=matrix(limit, limit); R[2, 1]=0; R[2, 2]=1; for(n=2, limit, print; print1("For n="n" in column n: "); Kappa_n=floor(n*log(3)/log(2)); a_n=0; for(k=n, Kappa_n, R[k+1, n]=R[k, n]+R[k, n-1]; print1(R[k+1, n]", "); a_n=a_n+R[k+1, n]); print; print(" and the sum is a(n)="a_n))} \\ Mike Winkler, Sep 12 2017
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CROSSREFS
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KEYWORD
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nonn,walk
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AUTHOR
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EXTENSIONS
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Two more terms from Jules Renucci (jules.renucci(AT)wanadoo.fr), Nov 02 2005
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STATUS
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approved
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