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 1-5 of 5 results.

A255495 2nd diagonal of triangle in A255494.

Original entry on oeis.org

1, 13, 130, 1106, 8575, 62475, 435576, 2939208, 19342285, 124800361, 792586270, 4969028750, 30822650251, 189500937303, 1156406300340, 7012380492516, 42294614785465, 253926386816725, 1518506730836026, 9050029200532298, 53778595325886295, 318762380704793571, 1885254096749834160
Offset: 0

Views

Author

N. J. A. Sloane, Mar 06 2015

Keywords

Crossrefs

Programs

  • Magma
    I:=[1,13,130,1106,8575,62475]; [n le 6 select I[n] else 14*Self(n-1) - 56*Self(n-2) +14*Self(n-3) +189*Self(n-4) + 84*Self(n-5) -20*Self(n-6): n in [1..31]]; // G. C. Greubel, Sep 20 2021
    
  • Mathematica
    a[n_]:= (1/2)*(Fibonacci[n+3, 2]*Fibonacci[n+4, 2] + 2^(n+4)*Fibonacci[n+4, 2] - 2*5^(n+3));
    Table[a[n], {n,0,30}] (* G. C. Greubel, Sep 20 2021 *)
  • Sage
    def P(n): return lucas_number1(n, 2, -1)
    def A255495(n): return (1/2)*(P(n+3)*P(n+4) + 2^(n+4)*P(n+4) - 2*5^(n+3))
    [A255495(n) for n in (0..30)] # G. C. Greubel, Sep 20 2021

Formula

G.f.: (1 -x +4*x^2)/((1+x)*(1-5*x)*(1-6*x+x^2)*(1-4*x-4*x^2)). - R. J. Mathar, Jun 14 2015
From G. C. Greubel, Sep 20 2021: (Start)
a(n) = (1/2)*(P(n+3)*P(n+4) + 2^(n+4)*P(n+4) - 2*5^(n+3)), where P(n) = A000129(n).
a(n) = 5*a(n-1) + P(n+1)*(P(n+3) - 2^(n+2)) = 5*a(n) + P(n+1)*A094706(n+1). (End)

Extensions

Terms a(13) onward from G. C. Greubel, Sep 20 2021

A255496 3rd diagonal of triangle in A255494.

Original entry on oeis.org

1, 38, 1106, 26544, 567203, 11179686, 207768576, 3692419776, 63361188037, 1057109514902, 17235551954894, 275697361933728, 4339725043253447, 67384965236252310, 1034147721558836220, 15711425790758327952, 236612932874975360809, 3536182524466029241958, 52494462902614684280330
Offset: 0

Views

Author

N. J. A. Sloane, Mar 06 2015

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_]:= (12)^(n+4) -(-2)^(n+1) -2^n*LucasL[2*n+9, 2] -5^(n+4)*Fibonacci[n+5, 2] +(1/10)*Fibonacci[n+4, 2]*(Fibonacci[n+4, 2]^2 +(-1)^n);
    Table[a[n], {n, 0, 30}] (* G. C. Greubel, Sep 20 2021 *)
  • Sage
    def P(n): return lucas_number1(n, 2, -1)
    def Q(n): return lucas_number2(n, 2, -1)
    def a(n): return (12)^(n+4) - (-2)^(n+1) - 2^n*Q(2*n+9) - 5^(n+4)*P(n+5) + (1/10)*P(n+4)*(P(n+4)^2 + (-1)^n)
    [a(n) for n in (0..30)] # G. C. Greubel, Sep 20 2021

Formula

From G. C. Greubel, Sep 20 2021: (Start)
a(n) = 12*a(n-1) + P(n+1)*A255495(n), where P(n) = A000129(n).
a(n) = (12)^(n+4) - (-2)^(n+1) - 2^n*Q(2*n+9) - 5^(n+4)*P(n+5) + (1/10)*P(n+4)*(P(n+4)^2 + (-1)^n), where P(n) = A000129(n), Q(n) = A002203(n).
G.f.: (1 -6*x +83*x^2 -228*x^3 -84*x^4 -200*x^5)/((1+2*x)*(1-12*x)*(1 +2*x -x^2)*(1 -10*x -25*x^2)*(1 -12*x +4*x^2)*(1 -14*x -x^2)). (End)

Extensions

3 more terms. - R. J. Mathar, Jun 14 2015
Terms a(12) onward added by G. C. Greubel, Sep 20 2021

A255497 4th diagonal of triangle in A255494.

Original entry on oeis.org

1, 105, 8575, 567203, 32897774, 1736613466, 85474679858, 3985272984490, 177983686766655, 7675333342669951, 321533970710475033, 13145650587005246037, 526435406695455725140, 20710119055883150135480, 802278112017623387734420, 30663507276425403310594244, 1158197029073059563909854477
Offset: 0

Views

Author

N. J. A. Sloane, Mar 06 2015

Keywords

Crossrefs

Programs

  • Mathematica
    A255496[n_]:= (12)^(n+4) -(-2)^(n+1) -2^n*LucasL[2*n+9, 2] -5^(n+4)*Fibonacci[n+5, 2] +(1/10)*Fibonacci[n+4, 2]*(Fibonacci[n+4, 2]^2 +(-1)^n);
    a[n_]:= a[n]= If[n<2, (105)^n, 29*a[n-1] + Fibonacci[n+1,2]*A255496[n]];
    Table[a[n], {n,0,30}] (* G. C. Greubel, Sep 20 2021 *)
  • Sage
    def P(n): return lucas_number1(n, 2, -1)
    def Q(n): return lucas_number2(n, 2, -1)
    def a(n): return (1/7680)*( 7680*(29)^(n+5) -192*(-5)^(n+6) -30 + Q(4*n+18) -96*5^(n+6)*Q(2*n+11) +12*(-1)^n*Q(2*n+9) +3*2^(n+10)*P(3*n+15) -640*(12)^(n+6)*P(n+6) -15*(-2)^(n+10)*P(n+5) )
    [a(n) for n in (0..30)] # G. C. Greubel, Sep 20 2021

Formula

From G. C. Greubel, Sep 20 2021: (Start)
a(n) = 29*a(n-1) + P(n+1)*A255496(n).
a(n) = (1/7680)*( 7680*(29)^(n+5) -192*(-5)^(n+6) -30 + Q(4*n+18) -96*5^(n+6)*Q(2*n+11) +12*(-1)^n*Q(2*n+9) +3*2^(n+10)*P(3*n+15) -640*(12)^(n+6)*P(n+6) -15*(-2)^(n+10)*P(n+5) ), where P(n) = A000129(n) and Q(n) = A002203(n).
G.f.: (1 -26*x +1108*x^2 -15042*x^3 +74319*x^4 +67340*x^5 +1376444*x^6 +2010720*x^7 -323920*x^8 +288000*x^9)/((1-x)*(1+5*x)*(1-29*x)*(1 +4*x -4*x^2)*(1 +6*x +x^2)*(1 -24*x -144*x^2)*(1 -28*x -4*x^2)*(1 -30*x +25*x^2)*(1 -34*x +x^2)).
(End)

Extensions

a(8)-a(10) from R. J. Mathar, Jun 14 2015
Terms a(11) onward added by G. C. Greubel, Sep 20 2021

A255498 5th diagonal of triangle in A255494.

Original entry on oeis.org

1, 280, 62475, 11179686, 1736613466, 243125885240, 31464032862802, 3828473678068060, 443307088929919375, 49283438913963499728, 5295767249826282145413, 552902424623732460251730, 56318224867097916236530640, 5615280578269206770801490160, 549533929275081475149009571700
Offset: 0

Views

Author

N. J. A. Sloane, Mar 06 2015

Keywords

Crossrefs

Programs

  • Mathematica
    P[n_]:= Fibonacci[n,2]; Q[n_]:= LucasL[n,2];
    A255497[n_]:= (1/7680)*(7680*(29)^(n+5) -192*(-5)^(n+6) -30 +Q[4*n+18] -96*5^(n+6)*Q[2*n+11] +12*(-1)^n*Q[2*n+9] +3*2^(n+10)*P[3*n+15] -640*(12)^(n+6)*P[n+6] -15*(-2)^(n+10)*P[n+5]);
    a[n_]:= a[n]= If[n<2, (280)^n, 70*a[n-1] +P[n+1]*A255497[n]];
    Table[a[n], {n, 0, 30}] (* G. C. Greubel, Sep 22 2021 *)
  • Sage
    @CachedFunction
    def P(n): return lucas_number1(n, 2, -1)
    def Q(n): return lucas_number2(n, 2, -1)
    def A255497(n): return (1/7680)*( 7680*(29)^(n+5) -192*(-5)^(n+6) -30 + Q(4*n+18) -96*5^(n+6)*Q(2*n+11) +12*(-1)^n*Q(2*n+9) +3*2^(n+10)*P(3*n+15) -640*(12)^(n+6)*P(n+6) -15*(-2)^(n+10)*P(n+5) )
    def a(n): return (280)^n if (n<2) else 70*a(n-1) + P(n+1)*A255497(n)
    [a(n) for n in (0..30)] # G. C. Greubel, Sep 20 2021

Formula

From G. C. Greubel, Sep 22 2021: (Start)
a(n) = 70*a(n-1) + A000129(n+1)*A255497(n), a(0) = 1, a(1) = 280.
a(n) = (1/222720)*(435*2^(n+7) + 2320*(-12)^(n+7) - 222720*(70)^(n+6) - 29*2^(n+6)*Q(4*n+22) + 1160*(12)^(n+7)*Q(2*n+13) + 87*(-2)^(n+8)*Q(2*n+11) +
P(5*n+25) - 2784*5^(n+6)*P(3*n+18) + 29*(-1)^n*P(3*n+15) + 7680*(29)^(n+7)*P(n + 7) + 2784*(-5)^(n+7)*P(n+6) - 174*P(n+5)), where P = A000129, Q(n) = A002203.
G.f.: (1 -96*x +11997*x^2 -596862*x^3 +15287055*x^4 -135141972*x^5 +366556867*x^6 -30606125134*x^7 - 254125754944*x^8 -657125309064*x^9 +376990806976*x^10 -2048614425760*x^11 +1171618742400*x^12 +77172576000*x^13 +29064960000*x^14)/((1-2*x)*(1+12*x)*(1-70*x)*(1 -2*x -x^2)*(1 +10*x -25*x^2)*(1 +12*x +4*x^2)*(1 +14*x -x^2)*(1 -58*x -841*x^2)*(1 -68*x +4*x^2)*(1 -70*x -25*x^2)*(1 -72*x +144*x^2)*(1 -82*x -x^2)). (End)

Extensions

Terms a(7) onward added by G. C. Greubel, Sep 22 2021

A094706 Convolution of Pell(n) and 2^n.

Original entry on oeis.org

0, 1, 4, 13, 38, 105, 280, 729, 1866, 4717, 11812, 29365, 72590, 178641, 438064, 1071153, 2613138, 6362965, 15470140, 37565389, 91125206, 220864377, 534951112, 1294960905, 3133261530, 7578261181, 18323338324, 44292046693, 107041649438
Offset: 0

Views

Author

Paul Barry, May 21 2004

Keywords

Crossrefs

Cf. A000079, A000129 (Pell numbers), A101164, A255494.

Programs

  • Magma
    I:=[0, 1, 4]; [n le 3 select I[n] else 4*Self(n-1)-3*Self(n-2)-2*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jun 24 2012
    
  • Mathematica
    LinearRecurrence[{4,-3, -2},{0,1,4},40] (* Vincenzo Librandi, Jun 24 2012 *)
  • Sage
    [lucas_number1(n+2, 2, -1) - 2^(n+1) for n in (0..30)] # G. C. Greubel, Sep 16 2021

Formula

G.f.: x/((1-2*x-x^2)*(1-2*x)).
a(n) = Sum_{k=0..n} ((1+sqrt(2))^n - (1-sqrt(2))^n)/(2*sqrt(2))*2^(n-k).
a(n) = (1 + 3*sqrt(2)/4)*(1 + sqrt(2))^n + (1 - 3*sqrt(2)/4)*(1-sqrt(2))^n - 2^(n+1).
a(n) = 4*a(n-1) - 3*a(n-2) - 2*a(n-3).
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k+1)*2^(n-2k-1);
a(n) = Sum_{k=0..n} binomial(k, n-k+1)*2^k*(1/2)^(n-k+1). - Paul Barry, Oct 07 2004
a(n) = sum of n-th row in A101164 = A000129(n) - A000079(n). - Reinhard Zumkeller, Dec 03 2004
a(n) = A000129(n+2) - 2^(n+1). - R. J. Mathar, Jan 29 2012
a(n) = 2*a(n-1) + A000129(n), with a(0) = 0, a(1) = 1. - G. C. Greubel, Sep 20 2021
Showing 1-5 of 5 results.