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A071734
a(n) = p(5n+4)/5 where p(k) denotes the k-th partition number.
18
1, 6, 27, 98, 315, 913, 2462, 6237, 15035, 34705, 77231, 166364, 348326, 710869, 1417900, 2769730, 5308732, 9999185, 18533944, 33845975, 60960273, 108389248, 190410133, 330733733, 568388100, 967054374, 1629808139, 2722189979
OFFSET
0,2
COMMENTS
One of the congruences related to the partition numbers stated by Ramanujan.
Also the coefficients in the expansion of C^5/B^6, in Watson's notation (p. 105). The connection to the partition function is in equation (3.31) with right side 5C^5/B^6 where B = x * f(-x^24), C = x^5 * f(-x^120) where f() is a Ramanujan theta function. Alternatively B = eta(q^24), C = eta(q^120). - Michael Somos, Jan 06 2015
REFERENCES
Berndt and Rankin, "Ramanujan: letters and commentaries", AMS-LMS, History of mathematics, vol. 9, pp. 192-193
G. H. Hardy, Ramanujan, Cambridge Univ. Press, 1940. - From N. J. A. Sloane, Jun 07 2012
LINKS
S. Bouroubi and N. Benyahia Tani, A new identity for complete Bell polynomials based on a formula of Ramanujan, J. Integer Seq. 12 (2009), 09.3.5.
J. L. Drost, A Shorter Proof of the Ramanujan Congruence Modulo 5, Amer. Math. Monthly 104(10) (1997), 963-964.
M. D. Hirschhorn, Another Shorter Proof of Ramanujan's Mod 5 Partition Congruence, and More, Amer. Math. Monthly 106(6) (1999), 580-583.
M. Savic, The Partition Function and Ramanujan's 5k+4 Congruence, Mathematics Exchange 1(1) (2003), 2-4.
G. N. Watson, Ramanujans Vermutung über Zerfällungszahlen, J. Reine Angew. Math. (Crelle) 179 (1938), 97-128.
Lasse Winquist, An elementary proof of p(11m+6) == 0 (mod 11), J. Combinatorial Theory 6(1) (1969), 56-59. MR0236136 (38 #4434). - From N. J. A. Sloane, Jun 07 2012
FORMULA
a(n) = (1/5)*A000041(5n+4).
G.f.: Product_{n>=1} (1 - x^(5*n))^5/(1 - x^n)^6 due to Ramanujan's identity. - Paul D. Hanna, May 22 2011
a(n) = A000041(A016897(n))/5 = A213260(n)/5. - Omar E. Pol, Jan 18 2013
Euler transform of period 5 sequence [ 6, 6, 6, 6, 1, ...]. - Michael Somos, Jan 07 2015
Expansion of q^(-19/24) * eta(q^5)^5 / eta(q)^6 in powers of q. - Michael Somos, Jan 07 2015
a(n) ~ exp(Pi*sqrt(10*n/3)) / (100*sqrt(3)*n). - Vaclav Kotesovec, Nov 28 2016
EXAMPLE
G.f. = 1 + 6*x + 27*x^2 + 98*x^3 + 315*x^4 + 913*x^5 + 2462*x^6 + ...
G.f. = q^19 + 6*q^43 + 27*q^67 + 98*q^91 + 315*q^115 + 913*q^139 + ...
MAPLE
with(combinat):
a:= n-> numbpart(5*n+4)/5:
seq(a(n), n=0..40); # Alois P. Heinz, Jan 07 2015
MATHEMATICA
a[ n_] := PartitionsP[ 5 n + 4] / 5; (* Michael Somos, Jan 07 2015 *)
a[ n_] := SeriesCoefficient[ 1 / QPochhammer[ x], {x, 0, 5 n + 4}] / 5; (* Michael Somos, Jan 07 2015 *)
nmax = 50; CoefficientList[Series[Product[(1 - x^(5*k))^5/(1 - x^k)^6, {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Nov 28 2016 *)
PROG
(PARI) {a(n) = if( n<0, 0, polcoeff( 1 / eta(x + O(x^(5*n + 5))), 5*n + 4) / 5)};
(PARI) {a(n) = numbpart(5*n + 4) / 5};
(PARI) a(n)=polcoeff(prod(m=1, n, (1-x^(5*m))^5/(1-x^m +x*O(x^n))^6), n) \\ Paul D. Hanna
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Benoit Cloitre, Jun 24 2002
STATUS
approved