A007317 -id:A007317 - cp's OEIS frontend

A126930 Inverse binomial transform of A005043.

1, -1, 2, -3, 6, -10, 20, -35, 70, -126, 252, -462, 924, -1716, 3432, -6435, 12870, -24310, 48620, -92378, 184756, -352716, 705432, -1352078, 2704156, -5200300, 10400600, -20058300, 40116600, -77558760, 155117520, -300540195, 601080390, -1166803110

Offset: 0

Philippe Deléham, Mar 17 2007

Successive binomial transforms are A005043, A000108, A007317, A064613, A104455. Hankel transform is A000012.
Moment sequence of the trace of the square of a random matrix in USp(2)=SU(2). If X=tr(A^2) is a random variable (a distributed with Haar measure) then a(n) = E[X^n]. - Andrew V. Sutherland, Feb 29 2008
From Tom Copeland, Nov 08 2014: (Start)
This array is one of a family of Catalan arrays related by compositions of the special fractional linear (Mobius) transformation P(x,t) = x/(1-t*x); its inverse Pinv(x,t) = P(x,-t); an o.g.f. of the Catalan numbers A000108, C(x) = [1-sqrt(1-4x)]/2; and its inverse Cinv(x) = x*(1-x). The Motzkin sums, or Riordan numbers, A005043 are generated by Mot(x)=C[P(x,1)]. One could, of course, choose the Riordan numbers as the parent sequence.
O.g.f.: G(x) = C[P[P(x,1),1]1] = C[P(x,2)] = (1-sqrt(1-4*x/(1+2*x)))/2 = x - x^2 + 2 x^3 - ... = Mot[P(x,1)].
Ginv(x) =  Pinv[Cinv(x),2] = P[Cinv(x),-2] = x(1-x)/[1-2x(1-x)] = (x-x^2)/[1-2(x-x^2)] = x*A146559(x).
Cf. A091867 and A210736 for an unsigned version with a leading 1. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Francesc Fite, Kiran S. Kedlaya, Victor Rotger and Andrew V. Sutherland, Sato-Tate distributions and Galois endomorphism modules in genus 2, arXiv preprint arXiv:1110.6638 [math.NT], 2011.
Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, L-polynomials and random matrices, arXiv:0803.4462 [math.NT], 2008-2010.
Paveł Szabłowski, Beta distributions whose moment sequences are related to integer sequences listed in the OEIS, Contrib. Disc. Math. (2024) Vol. 19, No. 4, 85-109. See p. 100.

Cf. A126120, A126869.

Cf. A000108, A005043, A146559, A091867, A210736.

egf := BesselI(0,2*x) - BesselI(1,2*x):
seq(n!*coeff(series(egf,x,34),x,n),n=0..33); # Peter Luschny, Dec 17 2014

CoefficientList[Series[(1 + 2 x - Sqrt[1 - 4 x^2])/(2 x (1 + 2 x)), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 23 2013 *)
Table[2^n Hypergeometric2F1[3/2, -n, 2, 2], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 02 2015 *)

x='x+O('x^50); Vec((1+2*x-sqrt(1-4*x^2))/(2*x*(1+2*x))) \\ Altug Alkan, Nov 03 2015

a(n) = (-1)^n*C(n, floor(n/2)) = (-1)^n*A001405(n).
a(2*n) = A000984(n), a(2*n+1) = -A001700(n).
a(n) = (1/Pi)*Integral_{t=0..Pi}(2cos(2t))^n*2sin^2(t) dt. - Andrew V. Sutherland, Feb 29 2008, Mar 09 2008
a(n) = (-2)^n + Sum_{k=0..n-1} a(k)*a(n-1-k), a(0)=1. - Philippe Deléham, Dec 12 2009
G.f.: (1+2*x-sqrt(1-4*x^2))/(2*x*(1+2*x)). - Philippe Deléham, Mar 01 2013
O.g.f.: (1 + x*c(x^2))/(1 + 2*x), with the o.g.f. c(x) for the Catalan numbers A000108. From the o.g.f. of the Riordan type Catalan triangle A053121. This is the rewritten g.f. given in the previous formula. This is G(-x) with the o.g.f. G(x) of A001405. - Wolfdieter Lang, Sep 22 2013
D-finite with recurrence (n+1)*a(n) +2*a(n-1) +4*(-n+1)*a(n-2)=0. - R. J. Mathar, Dec 04 2013
Recurrence (an alternative): (n+1)*a(n) = 8*(n-2)*a(n-3) + 4*(n-2)*a(n-2) + 2*(-n-1)*a(n-1), n>=3. - Fung Lam, Mar 22 2014
Asymptotics: a(n) ~ (-1)^n*2^(n+1/2)/sqrt(n*Pi). - Fung Lam, Mar 22 2014
E.g.f.: BesselI(0,2*x) - BesselI(1,2*x). - Peter Luschny, Dec 17 2014
a(n) = 2^n*hypergeom([3/2,-n], [2], 2). - Vladimir Reshetnikov, Nov 02 2015
G.f. A(x) satisfies: A(x) = 1/(1 + 2*x) + x*A(x)^2. - Ilya Gutkovskiy, Jul 10 2020

A349289 G.f. A(x) satisfies A(x) = 1 / ((1 - x) * (1 - x * A(x)^3)).

Original entry on oeis.org

1, 2, 9, 61, 493, 4371, 41065, 401563, 4044097, 41658044, 436862457, 4648331765, 50057856881, 544557984498, 5975422922413, 66059269445451, 735064865871889, 8226310738656892, 92531697191189777, 1045551973586825023, 11862334695799444993

Offset: 0

Ilya Gutkovskiy, Nov 13 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..500

Cf. A002293, A007317, A199475, A346646, A349290, A349291, A349292, A349293.

A349289 := proc(n)
    add( binomial(n+2*k,3*k)*binomial(4*k,k)/(3*k+1),k=0..n)  ;
end proc:
seq(A349289(n),n=0..50) ; # R. J. Mathar, Feb 10 2024

nmax = 20; A[] = 0; Do[A[x] = 1/((1 - x) (1 - x A[x]^3)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[Sum[Binomial[n + 2 k, 3 k] Binomial[4 k, k]/(3 k + 1), {k, 0, n}], {n, 0, 20}]

a(n) = Sum_{k=0..n} binomial(n+2*k,3*k) * binomial(4*k,k) / (3*k+1).
a(n) = F([1/4, 1/2, 3/4, (1+n)/2, (2+n)/2, -n], [1/3, 2/3, 2/3, 1, 4/3], -2^10/3^6) where F is the generalized hypergeometric function. - Stefano Spezia, Nov 13 2021
a(n) ~ sqrt(1 + 2*r) / (4 * 2^(1/6) * sqrt(3*Pi*(1-r)) * n^(3/2) * r^(n + 1/3)), where r = 0.0816785448577670972635343365300887975661075663022821172271... is the root of the equation 4^4 * r = 3^3 * (1-r)^3. - Vaclav Kotesovec, Nov 14 2021
D-finite with recurrence 81*n*(3*n-1)*(3*n+1)*a(n) +3*(243*n^3-7101*n^2+9986*n-3560)*a(n-1) +(-115027*n^3+514908*n^2-699869*n+269580)*a(n-2) +(-85543*n^3+1604715*n^2-6291692*n+6995280)*a(n-3) +(580211*n^3-6643158*n^2+23063299*n-23830944)*a(n-4) +(-33473*n^3-2231073*n^2+26352470*n-70945392)*a(n-5) +(-872129*n^3+17812344*n^2-119542699*n+264170868)*a(n-6) +(667171*n^3-14196243*n^2+100393472*n-236010000)*a(n-7) -6*(3*n-23)*(9948*n^2-147805*n+548868)*a(n-8) +4044*(3*n-26)*(n-8)*(3*n-22)*a(n-9)=0. - R. J. Mathar, Feb 10 2024
a(n) = 1 + Sum_{i, j, k, l>=0 and i+j+k+l=n-1} a(i) * a(j) * a(k) * a(l). - Seiichi Manyama, Jul 10 2025

A098474 Triangle read by rows, T(n,k) = C(n,k)C(2k,k)/(k+1), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 6, 5, 1, 4, 12, 20, 14, 1, 5, 20, 50, 70, 42, 1, 6, 30, 100, 210, 252, 132, 1, 7, 42, 175, 490, 882, 924, 429, 1, 8, 56, 280, 980, 2352, 3696, 3432, 1430, 1, 9, 72, 420, 1764, 5292, 11088, 15444, 12870, 4862, 1, 10, 90, 600, 2940, 10584, 27720

Offset: 0

Paul Barry, Sep 09 2004

A Catalan scaled binomial matrix.
From Philippe Deléham, Sep 01 2005: (Start)
Table U(n,k), k >= 0, n >= 0, read by antidiagonals, begins:
  row k = 0: 1, 1,  2,   5,   14, ... is A000108
  row k = 1: 1, 2,  6,  20,   70, ... is A000984
  row k = 2: 1, 3, 12,  50,  280, ... is A007854
  row k = 3: 1, 4, 20, 104,  548, ... is A076035
  row k = 4: 1, 5, 30, 185, 1150, ... is A076036
G.f. for row k: 1/(1-(k+1)*x*C(x)) where C(x) is the g.f. = for Catalan numbers A000108.
U(n,k) = Sum_{j=0..n} A106566(n,j)*(k+1)^j. (End)
This sequence gives the coefficients (increasing powers of x) of the Jensen polynomials for the Catalan sequence A000108 of degree n and shift 0. For the definition of Jensen polynomials for a sequence see a comment in A094436. - Wolfdieter Lang, Jun 25 2019

			Rows begin:
  1;
  1, 1;
  1, 2,  2;
  1, 3,  6,   5;
  1, 4, 12,  20,  14;
  1, 5, 20,  50,  70,  42;
  1, 6, 30, 100, 210, 252, 132;
  ...
Row 3: t*(1 - 3*t + 6*t^2 - 5*t^3)/(1 - 4*t)^(9/2) = 1/2*Sum_{k >= 1} k*(k+1)*(k+2)*(k+3)/4!*binomial(2*k,k)*t^k. - _Peter Bala_, Jun 13 2016

Indranil Ghosh, Rows 0..125, flattened

Row sums are A007317.
Antidiagonal sums are A090344.
Principal diagonal is A000108.
Mirror image of A124644.
Cf. A000984, A007854, A011973, A076035, A076036, A098443, A098473, A106566, A267633.

p := proc(n) option remember; if n = 0 then 1 else normal((x*(1 + 4*x)*diff(p(n-1, x), x) + (2*x + n + 1)*p(n-1, x))/(n + 1)) fi end:
row := n -> local k; seq(coeff(p(n), x, k), k = 0..n):
for n from 0 to 6 do row(n) od;  # Peter Luschny, Jun 21 2023

Table[Binomial[n, k] Binomial[2 k, k]/(k + 1), {n, 0, 10}, {k, 0, n}] // Flatten (* or *)
Table[(-1)^k*CatalanNumber[k] Pochhammer[-n, k]/k!, {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 17 2017 *)

from functools import cache
@cache
def A098474row(n: int) -> list[int]:
    if n == 0: return [1]
    a = A098474row(n - 1) + [0]
    row = [0] * (n + 1)
    row[0] = 1; row[1] = n
    for k in range(2, n + 1):
        row[k] = (a[k] * (n + k + 1) + a[k - 1] * (4 * k - 2)) // (n + 1)
    return row  # Peter Luschny, Jun 22 2023

def A098474(n,k):
    return (-1)^k*catalan_number(k)*rising_factorial(-n,k)/factorial(k)
for n in range(7): [A098474(n,k) for k in (0..n)] # Peter Luschny, Feb 05 2015

G.f.: 2/(1-x+(1-x-4*x*y)^(1/2)). - Vladeta Jovovic, Sep 11 2004
E.g.f.: exp(x*(1+2*y))*(BesselI(0, 2*x*y)-BesselI(1, 2*x*y)). - Vladeta Jovovic, Sep 11 2004
G.f.: 1/(1-x-xy/(1-xy/(1-x-xy/(1-xy/(1-x-xy/(1-xy/(1-x-xy/(1-xy/(1-... (continued fraction). - Paul Barry, Feb 11 2009
Sum_{k=0..n} T(n,k)*x^(n-k) = A126930(n), A005043(n), A000108(n), A007317(n+1), A064613(n), A104455(n) for x = -2, -1, 0, 1, 2, 3 respectively. - Philippe Deléham, Dec 12 2009
T(n,k) = (-1)^k*Catalan(k)*Pochhammer(-n,k)/k!. - Peter Luschny, Feb 05 2015
O.g.f.: [1 - sqrt(1-4tx/(1-x))]/(2tx) = 1 + (1+t) x + (1+2t+2t^2) x^2 + (1+3t+6t^2+5t^3) x^3 + ... , generating the polynomials of this entry, reverse of A124644. See A011973 for a derivation and the inverse o.g.f., connected to the Fibonacci, Chebyshev, and Motzkin polynomials. See also A267633. - Tom Copeland, Jan 25 2016
From Peter Bala, Jun 13 2016: (Start)
The o.g.f. F(x,t) = ( 1 - sqrt(1 - 4*t*x/(1 - x)) )/(2*t*x) satisfies the partial differential equation d/dx(x*(1 - x)*F) - x*t*(1 + 4*t)*dF/dt - 2*x*t*F = 1. This gives a recurrence for the row polynomials: (n + 2)*R(n+1,t) = t*(1 + 4*t)*R'(n,t) + (2*t + n + 2)*R(n,t), where the prime ' indicates differentiation with respect to t.
Equivalently, setting Q(n,t) = t^(n+2)*R(n,-t)/(1 - 4*t)^(n + 3/2) we have t^2*d/dt(Q(n,t)) = (n + 2)*Q(n+1,t).
This leads to the following expansions:
Q(0,t) = (1/2)*Sum_{k >= 1} k*binomial(2*k,k)*t^(k+1)
Q(1,t) = (1/2)*Sum_{k >= 1} k*(k+1)/2!*binomial(2*k,k)*t^(k+2)
Q(2,t) = (1/2)*Sum_{k >= 1} k*(k+1)*(k+2)/3!*binomial(2*k,k) *t^(k+3) and so on. (End)
Sum_{k=0..n} T(n,k)*x^k = A007317(n+1), A162326(n+1), A337167(n) for x = 1, 2, 3 respectively. - Sergii Voloshyn, Mar 31 2022

New name using a formula of Paul Barry by Peter Luschny, Feb 05 2015

A346648 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(6k,k) / (5k + 1).

Original entry on oeis.org

1, 2, 9, 73, 751, 8587, 104425, 1323952, 17303503, 231455104, 3153167249, 43597546197, 610232050453, 8629733401556, 123114479858631, 1769728635257503, 25607523627970183, 372688563309335806, 5451995469296025115, 80122698147986922194, 1182341393088427774071

Offset: 0

Ilya Gutkovskiy, Jul 26 2021

nonn

Binomial transform of A002295.

Seiichi Manyama, Table of n, a(n) for n = 0..500

Cf. A002295, A007317, A188687, A226910, A346646, A346647, A346649, A346650.

Table[Sum[Binomial[n, k] Binomial[6 k, k]/(5 k + 1), {k, 0, n}], {n, 0, 20}]
nmax = 20; A[] = 0; Do[A[x] = 1/(1 - x) + x (1 - x)^4 A[x]^6 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
nmax = 20; CoefficientList[Series[Sum[(Binomial[6 k, k]/(5 k + 1)) x^k/(1 - x)^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
Table[HypergeometricPFQ[{1/6, 1/3, 1/2, 2/3, 5/6, -n}, {2/5, 3/5, 4/5, 1, 6/5}, -46656/3125], {n, 0, 20}]

a(n) = sum(k=0, n, binomial(n,k)*binomial(6*k,k)/(5*k + 1)); \\ Michel Marcus, Jul 26 2021

G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x)^4 * A(x)^6.
G.f.: Sum_{k>=0} ( binomial(6*k,k) / (5*k + 1) ) * x^k / (1 - x)^(k+1).
a(n) ~ 49781^(n + 3/2) / (3359232 * sqrt(3*Pi) * n^(3/2) * 5^(5*n + 3/2)). - Vaclav Kotesovec, Jul 30 2021

A346649 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(7k,k) / (6k + 1).

Original entry on oeis.org

1, 2, 10, 95, 1146, 15343, 218407, 3241316, 49588850, 776483636, 12383420161, 200444399493, 3284531747403, 54378741581471, 908238222519566, 15284835717461020, 258933935458506210, 4412025177612412048, 75564998345532498844, 1300158755391113561288

Offset: 0

Ilya Gutkovskiy, Jul 26 2021

nonn

Binomial transform of A002296.

Seiichi Manyama, Table of n, a(n) for n = 0..500

Cf. A002296, A007317, A188687, A346646, A346647, A346648, A346650.

Table[Sum[Binomial[n, k] Binomial[7 k, k]/(6 k + 1), {k, 0, n}], {n, 0, 19}]
nmax = 19; A[] = 0; Do[A[x] = 1/(1 - x) + x (1 - x)^5 A[x]^7 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
nmax = 19; CoefficientList[Series[Sum[(Binomial[7 k, k]/(6 k + 1)) x^k/(1 - x)^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
Table[HypergeometricPFQ[{1/7, 2/7, 3/7, 4/7, 5/7, 6/7, -n}, {1/3, 1/2, 2/3, 5/6, 1, 7/6}, -823543/46656], {n, 0, 19}]

a(n) = sum(k=0, n, binomial(n,k)*binomial(7*k,k)/(6*k + 1)); \\ Michel Marcus, Jul 26 2021

G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x)^5 * A(x)^7.
G.f.: Sum_{k>=0} ( binomial(7*k,k) / (6*k + 1) ) * x^k / (1 - x)^(k+1).
a(n) ~ 870199^(n + 3/2) / (282475249 * sqrt(Pi) * n^(3/2) * 2^(6*n + 2) * 3^(6*n + 3/2)). - Vaclav Kotesovec, Jul 30 2021

A346650 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(8k,k) / (7k + 1).

Original entry on oeis.org

1, 2, 11, 120, 1661, 25484, 415619, 7066670, 123865313, 2222178999, 40604688117, 753051711707, 14138552326609, 268204210248763, 5132686807360949, 98973130183436759, 1921142366704203305, 37508070639707177792, 736080632477073862271, 14511777729474947626918

Offset: 0

Ilya Gutkovskiy, Jul 26 2021

nonn

Binomial transform of A007556.

In general, for m > 1, Sum_{k=0..n} binomial(n,k) * binomial(m*k,k) / ((m-1)*k + 1) ~ (m^m + (m-1)^(m-1))^(n + 3/2) / (sqrt(2*Pi) * m^((3*m-1)/2) * n^(3/2) * (m-1)^((m-1)*n + 3/2)). - Vaclav Kotesovec, Jul 30 2021

Seiichi Manyama, Table of n, a(n) for n = 0..500

Cf. A007317, A007556, A188687, A346646, A346647, A346648, A346649.

Table[Sum[Binomial[n, k] Binomial[8 k, k]/(7 k + 1), {k, 0, n}], {n, 0, 19}]
nmax = 19; A[] = 0; Do[A[x] = 1/(1 - x) + x (1 - x)^6 A[x]^8 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
nmax = 19; CoefficientList[Series[Sum[(Binomial[8 k, k]/(7 k + 1)) x^k/(1 - x)^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x]
Table[HypergeometricPFQ[{1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8, -n}, {2/7, 3/7, 4/7, 5/7, 6/7, 1, 8/7}, -16777216/823543], {n, 0, 19}]

a(n) = sum(k=0, n, binomial(n,k)*binomial(8*k,k)/(7*k + 1)); \\ Michel Marcus, Jul 26 2021

G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x)^6 * A(x)^8.
G.f.: Sum_{k>=0} ( binomial(8*k,k) / (7*k + 1) ) * x^k / (1 - x)^(k+1).
a(n) ~ 17600759^(n + 3/2) / (34359738368 * sqrt(Pi) * n^(3/2) * 7^(7*n + 3/2)). - Vaclav Kotesovec, Jul 30 2021

A277120 Number of branching factorizations of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 1, 5, 2, 3, 1, 11, 1, 3, 3, 15, 1, 11, 1, 11, 3, 3, 1, 45, 2, 3, 5, 11, 1, 19, 1, 51, 3, 3, 3, 62, 1, 3, 3, 45, 1, 19, 1, 11, 11, 3, 1, 195, 2, 11, 3, 11, 1, 45, 3, 45, 3, 3, 1, 113, 1, 3, 11, 188, 3, 19, 1, 11, 3, 19, 1, 345, 1, 3, 11, 11, 3

Offset: 1

Michel Marcus, Oct 01 2016

nonn

Per the formula, a(n) = 1 at prime n. As the sequence extends, additional values become more frequent than 1. These values can be characterized, for example, a(n) = 19 is seen at n corresponding to A007304, a(n) = 3 is seen at n corresponding to A006881, a(n) = 113 is seen at n corresponding to A085987. - Bill McEachen, Dec 28 2023
From Antti Karttunen, Jan 02 2024: (Start)
The value of a(n) depends only on the prime signature of n. In other words, for all i, j >= 1, it holds that A101296(i) = A101296(j) => a(i) = a(j). Moreover, it seems that the converse proposition also holds, that for all i, j >= 1, a(i) = a(j) => A101296(i) = A101296(j), i.e., for each distinct prime signature there exists a distinct value of a(n). This has been empirically checked up to the first 21001 prime signatures in A025487 (see A366884), and can be proved if one can show that the latter sequence (equally: A366377) is injective. If this conjecture holds, it would imply an unlimited number of statements like those given in the previous comment (see the formula section of A101296).
Questions: Are there any terms of the form 10k+4 or 10k+6? What is the asymptotic density of terms of the form 10k+5 (those ending with digit "5")? Compare to the data shown in A366884.
For squarefree n > 1, a(n) is never even, and apparently, never a multiple of five. See comments in A052886.
(End)

			In this scheme, the following factorizations of 12 are counted as distinct: 12, 2 x 6, 2 x (2 x 3), 2 x (3 x 2), 3 x 4, 3 x (2 x 2), 4 x 3, (2 x 2) x 3, 6 x 2, (2 x 3) x 2, (3 x 2) x 2, thus a(12) = 11. - _Antti Karttunen_, Nov 02 2016, based on the illustration given at page 14 of Knopfmacher & Mays paper.
The following factorizations of 30 are counted as distinct: 30, 2 x 15, 15 x 2, 3 x 10, 10 x 3, 5 x 6, 6 x 5, 2 x (3 x 5), 2 x (5 x 3), 3 x (2 x 5), 3 x (5 x 2), 5 x (2 x 3), 5 x (3 x 2), (2 x 3) x 5, (2 x 5) x 3, (3 x 2) x 5, (3 x 5) x 2, (5 x 2) x 3, (5 x 3) x 2, thus a(30) = 19. - _Antti Karttunen_, Jan 02 2024

Daniel Mondot, Table of n, a(n) for n = 1..9999
H.-K. Hwang, Théorèmes limites pour les structures combinatoires et les fonctions arithmétiques, PhD thesis, in Ecole Polytechnique, 1994. See page 198.
A. Knopfmacher, M. E. Mays, A survey of factorization counting functions, International Journal of Number Theory, 1(4):563-581,(2005). See page 14.
Index entries for sequences computed from exponents in factorization of n

Cf. A007317, A052886, A074206, A101296, A277130, A366377 [= a(A108951(n))], A366884 [= a(A025487(n))].

After n=1 differs from A104725 for the next time at n=32, where a(32) = 51, while A104725(32) = 52.

#include 
#define MAX 10000
/* Number of branching factorizations of n. */
unsigned long n, m, a, b, p, x, nbr[MAX];
int main(void)
{
  for (x=n=1; nDaniel Mondot, Oct 01 2016 */

v[n_] := v[n] = If[n == 1, 0, 1 + Sum[If[d == 1 || d^2 > n, 0, If[d^2 == n, 1, 2]*v[d]*v[n/d]], {d, Divisors[n]}]]; Table[v[n], {n, 1, 100}] (* Vaclav Kotesovec, Jan 13 2024, after Antti Karttunen *)

A277120(n) = if(1==n, 0, 1+sumdiv(n, d, if((1==d)||(d*d)>n,0,if((d*d)==n,1,2)*A277120(d)*A277120(n/d)))); \\ Antti Karttunen, Nov 02 2016, after Daniel Mondot's C-program above.

seq(n)={my(v=vector(n)); for(n=2, n, v[n] = 1 + sumdiv(n, d, v[d]*v[n/d])); v} \\ Andrew Howroyd, Nov 17 2018

a(1) = 0; for n > 1, a(n) = 1 + Sum_{d|n, 1 < d < n} a(d)*a(n/d). - Antti Karttunen, Nov 02 2016, after Daniel Mondot's C program, simplified Dec 30 2023.

For all n >= 1, a(prime^n) = A007317(n), and a(product of n distinct primes) = A052886(n). - Antti Karttunen, Dec 31 2023

More terms from Daniel Mondot, Oct 01 2016

A349290 G.f. A(x) satisfies A(x) = 1 / ((1 - x) * (1 - x * A(x)^4)).

Original entry on oeis.org

1, 2, 11, 96, 1001, 11456, 139013, 1756596, 22867421, 304560171, 4130200726, 56836946342, 791689962811, 11140615233281, 158140107648676, 2261708608884896, 32559326010349817, 471428798399646336, 6860801662510005266, 100302910051255600486

Offset: 0

Ilya Gutkovskiy, Nov 13 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..500

Cf. A002294, A007317, A199475, A346647, A349289, A349291, A349292, A349293.

nmax = 19; A[] = 0; Do[A[x] = 1/((1 - x) (1 - x A[x]^4)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[Sum[Binomial[n + 3 k, 4 k] Binomial[5 k, k]/(4 k + 1), {k, 0, n}], {n, 0, 19}]

a(n) = sum(k=0, n, binomial(n+3*k,4*k) * binomial(5*k,k) / (4*k+1)); \\ Michel Marcus, Nov 14 2021

a(n) = Sum_{k=0..n} binomial(n+3*k,4*k) * binomial(5*k,k) / (4*k+1).
a(n) = F([1/5, 2/5, 3/5, 4/5, (1+n)/3, (2+n)/3, (3+n)/3, -n], [1/4, 1/2, 1/2, 3/4, 3/4, 1, 5/4], -3^3*5^5/2^16), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 13 2021
a(n) ~ sqrt(1 + 3*r) / (2 * 5^(3/4) * sqrt(2*Pi*(1-r)) * n^(3/2) * r^(n + 1/4)), where r = 0.0631152861998150860738633360987635931... is the root of the equation 5^5 * r = 4^4 * (1-r)^4. - Vaclav Kotesovec, Nov 14 2021
a(n) = 1 + Sum_{i, j, k, l, m>=0 and i+j+k+l+m=n-1} a(i) * a(j) * a(k) * a(l) * a(m). - Seiichi Manyama, Jul 10 2025

A349291 G.f. A(x) satisfies A(x) = 1 / ((1 - x) * (1 - x * A(x)^5)).

Original entry on oeis.org

1, 2, 13, 139, 1775, 24886, 370099, 5733304, 91518691, 1494815215, 24862931821, 419674102147, 7170713484877, 123783319369420, 2155542171446485, 37820343323942566, 667957770644685811, 11865421405897931581, 211856917750711562695, 3800040255017879663415

Offset: 0

Ilya Gutkovskiy, Nov 13 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..500

Cf. A002295, A007317, A199475, A346648, A349289, A349290, A349292, A349293.

nmax = 19; A[] = 0; Do[A[x] = 1/((1 - x) (1 - x A[x]^5)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[Sum[Binomial[n + 4 k, 5 k] Binomial[6 k, k]/(5 k + 1), {k, 0, n}], {n, 0, 19}]

a(n) = Sum_{k=0..n} binomial(n+4*k,5*k) * binomial(6*k,k) / (5*k+1).
a(n) ~ sqrt(1 + 4*r) / (2^(6/5) * 3^(7/10) * sqrt(5*Pi*(1-r)) * n^(3/2) * r^(n + 1/5)), where r = 0.051436794119208432185504972091697516647... is the real root of the equation 6^6 * r = 5^5 * (1-r)^5. - Vaclav Kotesovec, Nov 14 2021
a(n) = 1 + Sum_{x_1, x_2, ..., x_6>=0 and x_1+x_2+...+x_6=n-1} Product_{k=1..6} a(x_k). - Seiichi Manyama, Jul 10 2025

A349292 G.f. A(x) satisfies A(x) = 1 / ((1 - x) * (1 - x * A(x)^6)).

Original entry on oeis.org

1, 2, 15, 190, 2871, 47643, 838888, 15389452, 290951545, 5629024955, 110908062511, 2217739684483, 44891645810124, 918086053852234, 18941156419798530, 393742848618632760, 8239112912485293357, 173406208518520952066, 3668419587671991125142

Offset: 0

Ilya Gutkovskiy, Nov 13 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..500

Cf. A002296, A007317, A199475, A346649, A349289, A349290, A349291, A349293.

nmax = 18; A[] = 0; Do[A[x] = 1/((1 - x) (1 - x A[x]^6)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[Sum[Binomial[n + 5 k, 6 k] Binomial[7 k, k]/(6 k + 1), {k, 0, n}], {n, 0, 18}]

a(n) = Sum_{k=0..n} binomial(n+5*k,6*k) * binomial(7*k,k) / (6*k+1).
a(n) ~ sqrt(1 + 5*r) / (2 * 7^(2/3) * sqrt(3*Pi*(1-r)) * n^(3/2) * r^(n + 1/6)), where r = 0.043408935906208378827553096713877784793679356... is the root of the equation 7^7 * r = 6^6 * (1-r)^6. - Vaclav Kotesovec, Nov 14 2021
a(n) = 1 + Sum_{x_1, x_2, ..., x_7>=0 and x_1+x_2+...+x_7=n-1} Product_{k=1..7} a(x_k). - Seiichi Manyama, Jul 11 2025

cp's OEIS Frontend

A126930 Inverse binomial transform of A005043.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A349289 G.f. A(x) satisfies A(x) = 1 / ((1 - x) * (1 - x * A(x)^3)).

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A098474 Triangle read by rows, T(n,k) = C(n,k)*C(2*k,k)/(k+1), n >= 0, 0 <= k <= n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Python

Sage

Formula

Extensions

A346648 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(6*k,k) / (5*k + 1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A346649 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(7*k,k) / (6*k + 1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A346650 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(8*k,k) / (7*k + 1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A277120 Number of branching factorizations of n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

A098474 Triangle read by rows, T(n,k) = C(n,k)C(2k,k)/(k+1), n >= 0, 0 <= k <= n.

A346648 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(6k,k) / (5k + 1).

A346649 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(7k,k) / (6k + 1).

A346650 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(8k,k) / (7k + 1).