A144796 -id:A144796 - cp's OEIS frontend

A144941 Numbers k such that 6*k-1 = A144796(k).

1, 36, 753, 41348, 868769, 47715364, 1002558481, 55063488516, 1156951618113, 63543218031908, 1335121164743729, 73328818545333124, 1540728667162644961, 84621393058096392996, 1777999546784527541073, 97653014260224692184068

Offset: 1

Richard Choulet, Sep 26 2008

Also the index of a pentagonal number which is equal to the sum of two consecutive pentagonal numbers. - Colin Barker, Dec 22 2014

			a(1) = 1 because 6*1 - 1 = 5 = A144796(1).

Colin Barker, Table of n, a(n) for n = 1..653
Index entries for linear recurrences with constant coefficients, signature (1,1154,-1154,-1,1).

Cf. A133301, A144796, A144797.

a:=[1,36,753,41348,868769];; for n in [6..30] do a[n]:=a[n-1] +1154*a[n-2]-1154*a[n-3]-a[n-4]+a[n-5]; od; a; # G. C. Greubel, Mar 16 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( x*(1+ 35*x-437*x^2+205*x^3+4*x^4)/((1-x)*(1-34*x+x^2)*(1+34*x+x^2)) )); // G. C. Greubel, Mar 16 2019

LinearRecurrence[{1,1154,-1154,-1,1},{1,36,753,41348,868769},30] (* Harvey P. Dale, Dec 27 2018 *)

Vec(-x*(1+35*x-437*x^2+205*x^3+4*x^4) / ((x-1)*(x^2-34*x+1)*(x^2+34*x+1)) + O(x^30)) \\ Colin Barker, Dec 22 2014

a=(x*(1+ 35*x-437*x^2+205*x^3+4*x^4)/((1-x)*(1-34*x+x^2)*(1+34*x +x^2))).series(x, 30).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Mar 16 2019

For the odd and even indices respectively the same recurrence is obtained: a(n+2) = 1154*a(n+1) - a(n) - 192.
We also have a(n+2) = 577*a(n+1) - 96 + 68*sqrt((72*a(n)^2-24*a(n)-32)).
G.f.: x*(1 + 35*x - 437*x^2 + 205*x^3 + 4*x^4) / ((1-x)*(1 - 34*x + x^2)*(1 + 34*x + x^2)). - R. J. Mathar, Nov 27 2011

a(6) corrected and sequence extended by R. J. Mathar, Nov 27 2011

A133301 a(n) is the n-th pentagonal number which is the sum of two consecutive pentagonal numbers.

Original entry on oeis.org

1, 1926, 850137, 2564464982, 1132138928657, 3415133918621062, 1507685261236261801, 4547981651299964079126, 2007805569980855008712097, 6056610836775865229750164742, 2673822786819976661810784866297, 8065673443881586606920210924732502

Offset: 1

Richard Choulet, Dec 20 2007

We solve the equation P(p) = P(r) + P(r+1) with unknowns p and r, equivalent to (6*p-1)^2 = 2*(6*r+2)^2 + 17. The Diophantine equation X^2 = 2*Y^2 + 17 whose solutions give p and r are obtained by (x(n), y(n)) such that:
x(1)=5, x(2)=215, x(3)=4517, x(4)=248087 and the same recurrence relation on the odd and even indices x(n+2) = 1154*x(n+1) - x(n);
y(1)=2, y(2)=152, y(3)=3194, y(4)=175424 and the same recurrence relation on the odd and even indices y(n+2) = 1154*y(n+1) - y(n).
The solutions (p,r) are given by the (u(n),v(n)) such that
u(1)=1, u(2)=36, u(3)=753, u(4)=41348 and the same recurrence relation on the odd and even indices u(n+2) = 1154*u(n+1) - u(n) - 192 or u(n+1) = 577*u(n) - 96 + 68*sqrt(72*u(n)^2 - 24*u(n) - 32);
v(1)=0, v(2)=25, v(3)=532, v(4)=29237 and the same recurrence relation on the odd and even indices v(n+2) = 1154*v(n+1) - v(n) + 384 or v(n+1) = 577*v(n) + 192 + 68*sqrt(72*u(n)^2 + 48*u(n) + 15).

			With P(m) = m*(3*m-1)/2,
a(1)=1 because a(1) = P(1) = P(0) + P(1);
a(2)=1926 because P(36) = 1926 = P(25) + P(26) = 925 + 1001;
a(3)=850137 because P(753) = 850137 = P(532) + P(533) = 424270 + 425867 ...

Colin Barker, Table of n, a(n) for n = 1..300
Index entries for linear recurrences with constant coefficients, signature (1,1331714,-1331714,-1,1).

Cf. A046173, A144796 (x(n)).

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/((1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2)) )); // G. C. Greubel, Mar 16 2019

a:=proc(m) if type (sqrt(18*m^2-6*m-8)/6-1/3), integer=true then m*(3*m-1)/2 else fi end : seq(a(m),m=1..100000)od; # Emeric Deutsch

# (3 # - 1)/2 &@ Select[Range[10^6], IntegerQ[Sqrt[18 #^2 - 6 # - 8]/6 - 1/3] &] (* or *)
Rest@ CoefficientList[Series[x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/( (1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2)), {x, 0, 12}], x] (* Michael De Vlieger, Jul 14 2016 *)

Vec(x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/((1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2)) + O(x^20)) \\ Colin Barker, Oct 20 2014

a=(x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/((1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2))).series(x, 20).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Mar 16 2019

For odd and even indices respectively: a(n+2) = 1331714*a(n+1) - a(n) - 416160; on the odd and the even indices respectively we also have a(n+1) = 665857*a(n) - 208080 + 19618*sqrt(1152*a(n)^2 - 720*a(n) - 32).

The g.f., h(z), is such that h(z) = a(1)*z + a(2)*z^2 + ... and is given by h(z) = z*(1 + 1925*z - 483503*z^2 + 65395*z^3 + 22*z^4)/((1-z)*(1 - 1331714*z^2 + z^4)).

Fixed typo in g.f. in formula, and more terms from Colin Barker, Oct 20 2014

A144942 Expansion of x^2(3x^3+145x^2-507x-25) / ((x-1)(x^2-34x+1)(x^2+34x+1)).

Original entry on oeis.org

0, 25, 532, 29237, 614312, 33739857, 708915900, 38935766125, 818088334672, 44931840368777, 944073229295972, 51851304849802917, 1089459688519217400, 59836360864832197825, 1257235536477947584012, 69051108586711506487517, 1450848719635862992732832

Offset: 1

Richard Choulet, Sep 26 2008

Also the index of the first of two consecutive pentagonal numbers whose sum is also a pentagonal number. - Colin Barker, Dec 22 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (1,1154,-1154,-1,1).

Cf. A133301, A144796, A144797.

CoefficientList[Series[x (3 x^3 + 145 x^2 - 507 x - 25)/((x - 1) (x^2 - 34 x + 1) (x^2 + 34 x + 1)), {x, 0, 20}], x] (* Vincenzo Librandi, Oct 20 2014 *)
LinearRecurrence[{1,1154,-1154,-1,1},{0,25,532,29237,614312},20] (* Harvey P. Dale, Jun 16 2025 *)

concat(0, Vec(x^2*(3*x^3+145*x^2-507*x-25)/((x-1)*(x^2-34*x+1)*(x^2+34*x+1)) + O(x^20))) \\ Colin Barker, Oct 20 2014

a(6) corrected, and more terms from Colin Barker, Oct 20 2014

Edited, name changed by Colin Barker, Oct 20 2014

cp's OEIS Frontend

A144941 Numbers k such that 6*k-1 = A144796(k).

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A133301 a(n) is the n-th pentagonal number which is the sum of two consecutive pentagonal numbers.

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A144942 Expansion of x^2*(3*x^3+145*x^2-507*x-25) / ((x-1)*(x^2-34*x+1)*(x^2+34*x+1)).

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A144942 Expansion of x^2(3x^3+145x^2-507x-25) / ((x-1)(x^2-34x+1)(x^2+34x+1)).