A027826 Inverse binomial transform of a_0 = 1, a_1, a_2, etc. is a_0, 0, a_1, 0, a_2, 0, etc.
1, 1, 2, 4, 9, 21, 50, 120, 290, 706, 1732, 4280, 10644, 26612, 66824, 168384, 425481, 1077529, 2733746, 6945812, 17669149, 44994345, 114682042, 292544200, 746831570, 1907983346, 4877966628, 12479883736, 31951158024, 81858610968, 209865391600, 538408691456
Offset: 0
Examples
Array of successive differences (col. 1 is the inverse binomial transform): 1, 1, 2, 4, 9, 21, 50, ... 0, 1, 2, 5, 12, 29, 70, ... 1, 1, 3, 7, 17, 41, ... 0, 2, 4, 10, 24, 59, ... 2, 2, 6, 14, 35, 87, ... 0, 4, 8, 21, 52, ... 4, 4, 13, 31, 79, ... 0, 9, 18, 48, ... 9, 9, 30, ... ...
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- N. J. A. Sloane, Transforms
Crossrefs
Cf. A051163.
Cf. A152193. - Gary W. Adamson, Nov 28 2008
Programs
-
Maple
a:= proc(n) option remember; add(`if`(k=0, 1, `if`(k::odd, 0, a(k/2)))*binomial(n, k), k=0..n) end: seq(a(n), n=0..40); # Alois P. Heinz, Jul 08 2015 -
Mathematica
a[n_] := a[n] = Sum[If[k == 0, 1, If[OddQ[k], 0, a[k/2]]]*Binomial[n, k], {k, 0, n}]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Jan 20 2017, translated from Maple *) -
PARI
a(n)=local(A,m); if(n<0,0,m=1; A=1+O(x); while(m<=n,m*=2; A=subst(A,x,(x/(1-x))^2)/(1-x)); polcoeff(A,n))
-
PARI
a=List();for(n=1,100,listput(a,sum(i=1,n\2,a[i]*binomial(n,2*i),1))) \\ M. F. Hasler, Aug 19 2015
Formula
G.f. A(x) satisfies A(x^2) = A(x/(1+x))/(1+x) and A(x) = A(x^2/(1-x)^2)/(1-x).
The recursive formula A[n+1] = A[n](x^2/(1-x)^2)/(1-x), A[0]=1, yields exactly 2^n terms after n iterations: A(x) - A[n](x) = x^(2^n) + (2^n+1)*x^(2^n+1) + O(x^(2^n+2)). For example, A[4] = (1-x)^3*(1-2*x-x^2)/((1-2*x)(1-4*x+4*x^2-2*x^4)) = A(x) - x^16 - 17*x^17 + O(x^18). - M. F. Hasler, Aug 19 2015
E.g.f.: exp(x) * Sum_{n>=0} a(n) * x^(2*n) / (2*n)!. - Ilya Gutkovskiy, Feb 26 2022
The expansion of exp(Sum_{n >= 1} a(n)*(2*x)^n/n!) = 1 + 2*x + 6*x^2 + 20*x^3 + 74*x^4 + 292*x^5 + 1204*x^6 + ... appears to have integer coefficients. Equivalently, the Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for positive integers k and n and all primes p >= 3. - Peter Bala, Jan 11 2023
Extensions
Incorrect g.f. and formulas removed by R. J. Mathar, Oct 02 2012
Incorrect g.f.s removed by Peter Bala, Jul 07 2015
Comments