A294833 -id:A294833 - cp's OEIS frontend

A005476 a(n) = n(5n - 1)/2.

0, 2, 9, 21, 38, 60, 87, 119, 156, 198, 245, 297, 354, 416, 483, 555, 632, 714, 801, 893, 990, 1092, 1199, 1311, 1428, 1550, 1677, 1809, 1946, 2088, 2235, 2387, 2544, 2706, 2873, 3045, 3222, 3404, 3591

Offset: 0

N. J. A. Sloane

a(n) is half the number of ways to divide an n X n square into 3 rectangles whose side-lengths are integers. See Matthew Scroggs link. - George Witty, Feb 06 2024

G. C. Greubel, Table of n, a(n) for n = 0..1000
Matthew Scroggs, Advent Calendar 2023 Solutions.
Leo Tavares, Illustration: Triangulated Diamonds
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A005475, A033994, A294833.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488.
Cf. similar sequences listed in A022288.
Cf. A016754, A133694, A000290.

List([0..40], n-> Binomial(5*n,2)/5); # G. C. Greubel, Jul 30 2019

[Binomial(5*n,2)/5: n in [0..40]]; // G. C. Greubel, Jul 30 2019

[seq(binomial(5*n,2)/5,n=0..40)]; # Zerinvary Lajos, Jan 02 2007

Table[n(5n-1)/2, {n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Oct 25 2008 *)

a(n)=n*(5*n-1)/2 \\ Charles R Greathouse IV, Sep 24 2015

[binomial(5*n,2)/5 for n in (0..40)] # G. C. Greubel, Jul 30 2019

a(n) = C(5*n,2)/5 for n>=0. - Zerinvary Lajos, Jan 02 2007
a(n) = A033991(n) - A000326(n). - Zerinvary Lajos, Jun 11 2007
a(n) = a(n-1) + 5*n - 3 for n>0, a(0)=0. - Vincenzo Librandi, Nov 18 2010
a(n) = A000217(n) + A000384(n) = A000290(n) + A000326(n). - Omar E. Pol, Jan 11 2013
a(n) = A130520(5*n+1). - Philippe Deléham, Mar 26 2013
a(n) = A033994(n) - A033994(n-1). - J. M. Bergot, Jun 12 2013
From Bruno Berselli, Oct 17 2016: (Start)
G.f.: x*(2 + 3*x)/(1 - x)^3.
a(n) = A000217(3*n-1) - A000217(2*n-1). (End)
E.g.f.: x*(4 + 5*x)*exp(x)/2. - G. C. Greubel, Jul 30 2019
Sum_{n>=1} 1/a(n) = 2 * A294833. - Amiram Eldar, Nov 16 2020
From Leo Tavares, Nov 20 2021: (Start)
a(n) = A016754(n) - A133694(n+1). See Triangulated Diamonds illustration.
a(n) = A000290(n) + A000217(n) + 2*A000217(n-1)
a(n) = 2*A000217(n) + 3*A000217(n-1). (End)

A294831 Numerators of the partial sums of the reciprocals of the numbers (k + 1)(5k + 4) = 2*A005476(k+1), for k >= 0.

Original entry on oeis.org

1, 11, 83, 410, 16799, 495151, 8516747, 55850623, 309309419, 1088610631, 6561497681, 777210281963, 12475578306953, 287734917200239, 10671842976127147, 844855135994501953, 846430303832665873, 75457260356268267017, 3551759427031132995079, 711302288219532928235, 712163917143684270659

Offset: 0

Wolfdieter Lang, Nov 18 2017

The corresponding denominators are given in A294832.
For the general case V(m,r;n) = Sum_{k=0..n} 1/((k + 1)*(m*k + r)) = (1/(m - r))*Sum_{k=0..n} (m/(m*k + r) - 1/(k+1)), for r = 1, ..., m-1 and m = 2, 3, ..., and their limits see a comment in A294512. Here [m,r] = [5,4].
The limit of the series is V(5,4) = lim_{n -> oo} V(5,4;n) = ((5/2)*log(5) + (2*phi - 1)*(log(phi) - (Pi/5)*sqrt(3 + 4*phi)))/2, with the golden section phi:= (1 + sqrt(5))/2 = A001622. The value is 0.3877929018046... given in A294833.
In the Koecher reference v_5(4) = (1/5)*V(5,4) = 0.07755858036... given there by (1/4)*log(5) + (1/(2*sqrt(5)))*log((1 + sqrt(5))/2) - (Pi/10)*sqrt((5 + 2*sqrt(5))/5).

			The rationals V(5,4;n), n >= 0, begin:1/4, 11/36, 83/252, 410/1197, 16799/47880, 495151/1388520, 8516747/23604840, 55850623/153431460, 309309419/843873030, 1088610631/2953555605, 6561497681/17721333630, 777210281963/2091117368340, 12475578306953/33457877893440, ...
V(5,4;10^6) = 0.3877927018 (Maple 10 digits) to be compared with 0.3877929018 obtained from A294833 with 10 digits.

Max Koecher, Klassische elementare Analysis, Birkhäuser, Basel, Boston, 1987, Eulersche Reihen, pp. 189-193. For v_5(4) see p. 192.

G. C. Greubel, Table of n, a(n) for n = 0..600
Eric Weisstein's World of Mathematics, Digamma Function

Cf. A001622, 2*A005476, A294512, A294828/A294829 (V(5,3;n)), A294832, A294833.

[Numerator((&+[1/((k + 1)*(5*k + 4)): k in [0..n]])): n in [0..50]]; // G. C. Greubel, Aug 30 2018

Table[Numerator[Sum[1/((k + 1)*(5*k + 4)), {k, 0, n}]], {n, 0, 50}] (* G. C. Greubel, Aug 30 2018 *)

a(n) = numerator(sum(k=0, n, 1/((k + 1)*(5*k + 4)))); \\ Michel Marcus, Nov 19 2017

a(n) = numerator(V(5,4;n)) with V(5,4;n) = Sum_{k=0..n} 1/((k + 1)*(5*k + 4)) = Sum_{k=0..n} 1/(2*A005476(k+1)) = Sum_{k=0..n} (1/(k + 4/5) - 1/(k+1)) = -Psi(4/5) + Psi(n+9/5) - (gamma + Psi(n+2)) with the digamma function Psi and the Euler-Mascheroni constant gamma = -Psi(1) from A001620.

cp's OEIS Frontend

A005476 a(n) = n(5n - 1)/2.

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A294831 Numerators of the partial sums of the reciprocals of the numbers (k + 1)(5k + 4) = 2*A005476(k+1), for k >= 0.

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cp's OEIS Frontend

A005476 a(n) = n*(5*n - 1)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

A294831 Numerators of the partial sums of the reciprocals of the numbers (k + 1)*(5*k + 4) = 2*A005476(k+1), for k >= 0.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A005476 a(n) = n(5n - 1)/2.

A294831 Numerators of the partial sums of the reciprocals of the numbers (k + 1)(5k + 4) = 2*A005476(k+1), for k >= 0.