cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 141 results. Next

A260793 Primes p such that p does not divide any term of the Apéry-like sequence A000172 (also known as Type I primes).

Original entry on oeis.org

3, 11, 17, 19, 43, 83, 89, 97, 113, 137, 139, 163, 193, 211, 233, 241, 283, 307, 313, 331, 347, 353, 379, 401, 409, 419, 433, 443, 491, 499, 523, 547, 569, 587, 601, 617, 619, 641, 643, 673, 811, 827, 859, 881, 929, 947, 953, 977, 1009, 1019, 1033, 1049, 1051
Offset: 1

Views

Author

N. J. A. Sloane, Aug 05 2015

Keywords

Comments

See Schulte et al. (2014) for the precise definition of Type I primes.

Crossrefs

For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see this sequence, A291275-A291284 and A133370 respectively.

Programs

  • Mathematica
    maxPrime = 1051;
    maxPi = PrimePi @ maxPrime;
    okQ[p_] := AllTrue[Range[3 maxPi (* coeff 3 is empirical *)], GCD[HypergeometricPFQ[{-#, -#, -#}, {1, 1}, -1], p] == 1&];
    Select[Prime[Range[maxPi]], okQ] (* Jean-François Alcover, Jan 13 2020 *)

Extensions

Edited by N. J. A. Sloane, Aug 22 2017

A166990 G.f.: A(x) = exp( Sum_{n>=1} A000172(n)*x^n/n ) where Franel number A000172(n) = Sum_{k=0..n} C(n,k)^3.

Original entry on oeis.org

1, 2, 7, 30, 147, 786, 4472, 26644, 164477, 1044258, 6782484, 44887236, 301782361, 2056250570, 14172792355, 98667874038, 692948001906, 4904403499992, 34951124337300, 250617829087656, 1807055528439771, 13095146839953030
Offset: 0

Views

Author

Paul D. Hanna, Nov 17 2009

Keywords

Comments

Analogous to the square of the g.f. of Catalan numbers (A000108):
C(x)^2 = exp( Sum_{n>=1} A000984(n)*x^n/n ) where central binomial coefficient A000984(n) = Sum_{k=0..n} C(n,k)^2.

Examples

			G.f.: A(x) = 1 + 2*x + 7*x^2 + 30*x^3 + 147*x^4 + 786*x^5 + 4472*x^6 +...
log(A(x)) = 2*x + 10*x^2/2 + 56*x^3/3 + 346*x^4/4 + 2252*x^5/5 + 15184*x^6/6 + 104960*x^7/7 +...+ A000172(n)*x^n/n +...
		

Crossrefs

Cf. A000172 (Franel numbers), A166991, A166992, A218117, A218119.

Programs

  • Mathematica
    a[n_] := Sum[(Binomial[n, k])^3, {k, 0, n}]; f[x_] := Sum[a[n]*x^n/n, {n, 1, 75}]; CoefficientList[Series[Exp[f[x]], {x, 0, 50}], x] (* G. C. Greubel, May 30 2016 *)
    nmax = 30; Clear[a]; franel = RecurrenceTable[{n^2*a[n] == (7*n^2 - 7*n + 2)*a[n-1] + 8*(n-1)^2*a[n-2], a[1] == 2, a[2] == 10}, a, {n, 1, nmax}]; $RecursionLimit -> Infinity; a[n_] := a[n] = If[n == 0, 1, Sum[franel[[k]]*a[n-k], {k, 1, n}]/n]; Table[a[n], {n, 0, nmax}] (* Vaclav Kotesovec, Oct 27 2024 *)
  • PARI
    {a(n)=polcoeff(exp(sum(m=1,n,sum(k=0,m,binomial(m,k)^3)*x^m/m)+x*O(x^n)),n)}

Formula

Self-convolution of A166991.
a(n) ~ c * 8^n / n^2, where c = 0.58462945... - Vaclav Kotesovec, Nov 27 2017, updated Oct 29 2024

A242169 Least prime divisor of Fr(n) which does not divide any Fr(k) with k < n, or 1 if such a primitive prime divisor of Fr(n) does not exist, where Fr(n) denotes the n-th Franel number given by A000172.

Original entry on oeis.org

2, 5, 7, 173, 563, 13, 41, 369581, 937, 61, 23, 29, 2141, 12148537, 31, 157, 59, 37, 506251, 151, 3019, 769, 47, 6730949, 79, 53, 3853, 661, 138961158000728258971, 1361, 421, 96920594213, 51378681049, 457, 71
Offset: 1

Views

Author

Zhi-Wei Sun, May 05 2014

Keywords

Comments

Conjecture: a(n) > 1 for all n > 0.

Examples

			a(7) = 41 since Fr(7) = 2^9*5*41 with the prime factor 41 dividing none of Fr(1), ..., Fr(6) but 2 divides Fr(1) = 2 and 5 divides Fr(2) = 10.
		

Crossrefs

Programs

  • Mathematica
    Fr[n_]:=Sum[Binomial[n,k]^3,{k,0,n}]
    f[n_]:=FactorInteger[Fr[n]]
    p[n_]:=Table[Part[Part[f[n],k],1],{k,1,Length[f[n]]}]
    Do[Do[Do[If[Mod[Fr[i],Part[p[n],k]]==0,Goto[aa]],{i,1,n-1}];Print[n," ",Part[p[n],k]];Goto[bb];Label[aa];Continue,{k,1,Length[p[n]]}];Print[n," ",1];Label[bb];Continue,{n,1,35}]

A232687 G.f. A(x) satisfies: the sum of the coefficients of x^k, k=0..n, in A(x)^n equals Sum_{k=0..n} C(n,k)^3 = A000172(n) (Franel numbers), for n>=0.

Original entry on oeis.org

1, 1, 3, 7, 20, 66, 244, 980, 4182, 18674, 86353, 410541, 1996214, 9888844, 49760925, 253767097, 1309154825, 6822023553, 35865392690, 190038440422, 1014015337209, 5444707218851, 29401289997403, 159584901816255, 870267544114291, 4766246752344215, 26206635040151511
Offset: 0

Views

Author

Paul D. Hanna, Dec 05 2013

Keywords

Comments

Compare to: Sum_{k=0..n} [x^k] 1/(1-x)^n = Sum_{k=0..n} C(n,k)^2 = (2*n)!/n!^2.
a(n+1)/a(n) tends to 6.0295... - Vaclav Kotesovec, Jan 22 2014

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 7*x^3 + 20*x^4 + 66*x^5 + 244*x^6 + 980*x^7 +...
ILLUSTRATION OF INITIAL TERMS.
If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
A^0: [1],0,  0,   0,    0,    0,     0,      0,      0, ...;
A^1: [1, 1], 3,   7,   20,   66,   244,    980,   4182, ...;
A^2: [1, 2,  7], 20,   63,  214,   789,   3124,  13112, ...;
A^3: [1, 3, 12,  40], 138,  492,  1848,   7326,  30531, ...;
A^4: [1, 4, 18,  68,  255], 960,  3716,  14920,  62295, ...;
A^5: [1, 5, 25, 105,  425, 1691], 6785,  27805, 117165, ...;
A^6: [1, 6, 33, 152,  660, 2772, 11560], 48588, 207774, ...;
A^7: [1, 7, 42, 210,  973, 4305, 18676,  80746],351792, ...;
A^8: [1, 8, 52, 280, 1378, 6408, 28916, 128808, 573311], ...; ...
then the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals Sum_{k=0..n} C(n,k)^3 = A000172(n):
A000172(0) = 1 = 1;
A000172(1) = 1 + 1 = 2;
A000172(2) = 1 + 2 +  7 = 10;
A000172(3) = 1 + 3 + 12 +  40 = 56;
A000172(4) = 1 + 4 + 18 +  68 + 255 = 346;
A000172(5) = 1 + 5 + 25 + 105 + 425 + 1691 = 2252;
A000172(6) = 1 + 6 + 33 + 152 + 660 + 2772 + 11560 = 15184; ...
		

Crossrefs

Programs

  • Mathematica
    Franel[n_] := Sum[Binomial[n, k]^3, {k, 0, n}];
    a[0] = 1; a[n_] := Module[{B, G}, B = Sum[Franel[k]*x^k, {k, 0, n+1}] + x^3*O[x]^n; G = 1+x*O[x]^n; For[i=1, i <= n, i++, G = 1+Integrate[(B-1)* (G/x)-B*G^2, x]]; SeriesCoefficient[x/InverseSeries[x*G, x], {x, 0, n}]];
    Table[a[n], {n, 0, 26}] (* Jean-François Alcover, Jan 15 2018, translated from 2nd PARI program *)
  • PARI
    /* By Definition (slow): */
    {Franel(n)=sum(k=0,n,binomial(n,k)^3)}
    {a(n)=if(n==0, 1, (Franel(n) - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j)^n + x*O(x^k), k)))/n)}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* Faster, using series reversion: */
    {Franel(n)=sum(k=0,n,binomial(n,k)^3)}
    {a(n)=local(B=sum(k=0, n+1, Franel(k)*x^k)+x^3*O(x^n), G=1+x*O(x^n));
    for(i=1, n, G = 1 + intformal( (B-1)*G/x - B*G^2)); polcoeff(x/serreverse(x*G), n)}
    for(n=0, 30, print1(a(n), ", "))

Formula

Given g.f. A(x), Sum_{k=0..n} [x^k] A(x)^n = Sum_{k=0..n} C(n,k)^3 = A000172(n).
Given g.f. A(x), let G(x) = A(x*G(x)) then (G(x) + x*G'(x)) / (G(x) - x*G(x)^2) = Sum_{n>=0} (3*n)!/n!^3 * x^(2*n)/(1-2*x)^(3*n+1) = Sum_{n>=0} A000172(n)*x^n.

A166991 G.f.: A(x) = exp( Sum_{n>=1} A000172(n)*x^n/(2*n) ) where Franel number A000172(n) = Sum_{k=0..n} C(n,k)^3.

Original entry on oeis.org

1, 1, 3, 12, 57, 300, 1693, 10045, 61890, 392688, 2550843, 16891566, 113660475, 775223595, 5349057132, 37280705406, 262119009927, 1857241951359, 13250054817027, 95110710932424, 686490953423700, 4979704242810870
Offset: 0

Views

Author

Paul D. Hanna, Nov 17 2009

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 12*x^3 + 57*x^4 + 300*x^5 + 1693*x^6 +...
log(A(x)^2) = 2*x + 10*x^2/2 + 56*x^3/3 + 346*x^4/4 + 2252*x^5/5 + 15184*x^6/6 + 104960*x^7/7 +...+ A000172(n)*x^n/n +...
		

Crossrefs

Cf. A000172 (Franel numbers), A166990, A166993, A218118, A218120.

Programs

  • Mathematica
    a[n_] := Sum[(Binomial[n, k])^3, {k, 0, n}]; f[x_] := Sum[a[n]*x^n/(2*n), {n, 1, 75}]; CoefficientList[Series[Exp[f[x]], {x, 0, 50}], x] (* G. C. Greubel, May 30 2016 *)
  • PARI
    {a(n)=polcoeff(exp(sum(m=1,n,sum(k=0,m,binomial(m,k)^3)/2*x^m/m)+x*O(x^n)),n)}

Formula

Self-convolution yields A166990.
a(n) ~ c * 8^n / n^2, where c = 0.231776... - Vaclav Kotesovec, Nov 27 2017

A248139 Least positive integer m such that m + n divides f(m) + f(n), where f(.) is given by A000172.

Original entry on oeis.org

1, 1, 25, 6, 14, 4, 13, 49, 19, 10, 2, 56, 2, 5, 6, 5, 27, 61, 9, 33, 23, 53, 21, 15, 3, 24, 11, 58, 39, 118, 3, 1598, 20, 40, 4, 2, 58, 26, 29, 17, 47, 34, 4, 31, 43, 163, 41, 25, 8, 26, 67, 40, 21, 214, 535, 12, 7, 22, 164, 74
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 02 2014

Keywords

Comments

Conjecture: a(n) exists for any n > 0.

Examples

			a(5) = 14 since 5 + 14 = 19 divides f(5) + f(14) = 2252 + 112738423360 = 112738425612 = 19*5933601348.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=Sum[Binomial[n,k]^3,{k,0,n}]
    Do[m=1; Label[aa]; If[Mod[f[m]+f[n], m+n]==0, Print[n, " ", m]; Goto[bb]]; m=m+1; Goto[aa]; Label[bb]; Continue, {n, 1, 60}]

A216354 G.f.: A(x) = exp( Sum_{n>=1} A000172(n)^n*x^n/n ) where Franel number A000172(n) = Sum_{k=0..n} C(n,k)^3.

Original entry on oeis.org

1, 2, 52, 58640, 3583098592, 11584364000042912, 2042518153012624794424576, 20047892010468651075834167466942080, 11138509206681372983092694151616405935206616064, 354938139483847646086359348765071470756626699510545192807936
Offset: 0

Views

Author

Paul D. Hanna, Sep 04 2012

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 52*x^2 + 58640*x^3 + 3583098592*x^4 +...
where
log(A(x)) = 2*x + 10^2*x^2/2 + 56^3*x^3/3 + 346^4*x^4/4 + 2252^5*x^5/5 + 15184^6*x^6/6 + 104960^7*x^7/7 +...+ A000172(n)^n*x^n/n +...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^3)^m*x^m/m+x*O(x^n)))); polcoeff(A, n)}
    for(n=0, 31, print1(a(n), ", "))

A052144 a(n) = A000172(n)^2.

Original entry on oeis.org

1, 4, 100, 3136, 119716, 5071504, 230553856, 11016601600, 546360462244, 27888242788624, 1456587070867600, 77515424509446400, 4189899499315360000, 229472379264509977600, 12709952101698593689600, 710863065714510068187136
Offset: 0

Views

Author

N. J. A. Sloane, Jan 23 2000

Keywords

References

  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see page 191.

Crossrefs

Cf. A000172.

Programs

Formula

P-recursive: P(n-1)*n^4*a(n) = P(n)*Q(n)*a(n-1) + 8*P(n-1)*Q(n)*a(n-2) - 512*P(n)*(n-2)^4*a(n-3), where P(n) = 7*n^2 - 7*n + 2 and Q(n) = 57*n^4 - 228*n^3 + 321*n^2 - 186*n + 40 with a(0) = 1, a(1) = 4 and a(2) = 100. - Peter Bala, Feb 01 2024
a(n) ~ 2^(6*n+2) / (3*Pi^2*n^2). - Vaclav Kotesovec, Feb 02 2024

A181418 a(n) = A000984(n)*A000172(n), which is the term-wise product of the Central binomial coefficients and Franel numbers, respectively.

Original entry on oeis.org

1, 4, 60, 1120, 24220, 567504, 14030016, 360222720, 9513014940, 256758913840, 7051260776560, 196403499277440, 5535202897806400, 157551884911456000, 4522682234563776000, 130783762623673221120, 3806221127760278029980
Offset: 0

Views

Author

Paul D. Hanna, Jan 28 2011

Keywords

Comments

This sequence is s_6 in Cooper's paper. - Jason Kimberley, Nov 25 2012
Diagonal of the rational function R(x,y,z,w)=1/(1-(w*x*y+w*z+x*y+x*z+y+z)). - Gheorghe Coserea, Jul 13 2016

Examples

			E.g.f.: A(x) = 1 + 4*x/2! + 60*x^2/(2!*4!) + 1120*x^3/(3!*6!) + 24220*x^4/(4!*8!) + 567504*x^5/(5!*10!) +....
where A(x)^(1/2) = 1 + x + x^2/2!^3 + x^3/3!^3 + x^4/4!^3 +x^5/5!^3 +...
		

Crossrefs

Related to diagonal of rational functions: A268545-A268555.

Programs

  • Mathematica
    Table[Binomial[2n,n]*Sum[Binomial[n,k]^3,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Mar 06 2014 *)
  • PARI
    {a(n)=binomial(2*n,n)*sum(k=0,n,binomial(n,k)^3)}
    
  • PARI
    {a(n)=(2*n)!*n!*polcoeff(sum(m=0, n, x^m/m!^3+x*O(x^n))^2, n)}

Formula

a(n) = C(2n,n) * Sum_{k=0..n} C(n,k)^3.
E.g.f.: Sum_{n>=0} a(n)*x^n/(n!*(2*n)!) = ( Sum_{n>=0} x^n/n!^3 )^2.
From Jason Kimberley, Nov 26 2012: (Start)
1/Pi
= (2/25)*Sum_{n>=0} a(n)*(9n+2)/50^n. [Cooper, equation (5)]
= (2/25)*Sum_{n>=0} a(n)*A017185(n)/A165800(n). (End)
G.f.: 4*hypergeom([1/6, 1/3],[1],(27/2)*(1+(1-32*x)^(1/2))*(1-(1-32*x)^(1/2))^2/(3+(1-32*x)^(1/2))^3)^2/(3+(1-32*x)^(1/2)). - Mark van Hoeij, May 07 2013
Recurrence: n^3*a(n) = 2*(2*n-1)*(7*n^2 - 7*n + 2)*a(n-1) + 32*(n-1)*(2*n-3)*(2*n-1)*a(n-2). - Vaclav Kotesovec, Mar 06 2014
a(n) ~ 2^(5*n+1) / (sqrt(3) * (Pi*n)^(3/2)). - Vaclav Kotesovec, Mar 06 2014
0 = (-x^2+28*x^3+128*x^4)*y''' + (-3*x+126*x^2+768*x^3)*y'' + (-1+92*x+864*x^2)*y' + (4+96*x)*y, where y is g.f. - Gheorghe Coserea, Jul 13 2016

A216352 G.f.: A(x) = exp( Sum_{n>=1} A000172(n)^2*x^n/n ) where Franel number A000172(n) = Sum_{k=0..n} C(n,k)^3.

Original entry on oeis.org

1, 4, 58, 1256, 35771, 1200188, 45016678, 1827941560, 78753548245, 3551810922324, 166120394053698, 8002733850225288, 395089619067741926, 19911864121386482264, 1021345223473335336668, 53190166903606336969840, 2807000233813092463820488, 149869216802426305919295328
Offset: 0

Views

Author

Paul D. Hanna, Sep 04 2012

Keywords

Examples

			G.f.: A(x) = 1 + 4*x + 58*x^2 + 1256*x^3 + 35771*x^4 + 1200188*x^5 +...
such that
log(A(x)) = 4*x + 100*x^2/2 + 3136*x^3/3 + 119716*x^4/4 + 5071504*x^5/5 +...+ A000172(n)^2*x^n/n +...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^3)^2*x^m*1^m/m+x*O(x^n)))); polcoeff(A, n)}
    for(n=0, 31, print1(a(n), ", "))
Showing 1-10 of 141 results. Next