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 1-6 of 6 results.

A132615 Triangle T, read by rows, where row n+1 of T = row n of T^(2n-1) with appended '1' for n>=0 with T(0,0)=1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 6, 3, 1, 1, 80, 25, 5, 1, 1, 1666, 378, 56, 7, 1, 1, 47232, 8460, 1020, 99, 9, 1, 1, 1694704, 252087, 26015, 2134, 154, 11, 1, 1, 73552752, 9392890, 855478, 61919, 3848, 221, 13, 1, 1, 3744491970, 420142350, 34461260, 2257413, 125760, 6290, 300, 15, 1, 1
Offset: 0

Views

Author

Paul D. Hanna, Aug 24 2007

Keywords

Examples

			Triangle begins:
  1;
  1, 1;
  1, 1, 1;
  6, 3, 1, 1;
  80, 25, 5, 1, 1;
  1666, 378, 56, 7, 1, 1;
  47232, 8460, 1020, 99, 9, 1, 1;
  1694704, 252087, 26015, 2134, 154, 11, 1, 1;
  73552752, 9392890, 855478, 61919, 3848, 221, 13, 1, 1; ...
GENERATE T FROM ODD MATRIX POWERS OF T.
Matrix cube, T^3, begins:
  1;
  3, 1;
  6, 3, 1; <-- row 3 of T
  31, 12, 3, 1;
  357, 100, 18, 3, 1;
  6786, 1455, 205, 24, 3, 1; ...
where row 3 of T = row 2 of T^3 with appended '1'.
Matrix fifth power, T^5, begins:
  1;
  5, 1;
  15, 5, 1;
  80, 25, 5, 1; <-- row 4 of T
  855, 215, 35, 5, 1;
  15171, 3065, 410, 45, 5, 1; ...
where row 4 of T = row 3 of T^5 with appended '1'.
Matrix seventh power, T^7, begins:
  1;
  7, 1;
  28, 7, 1;
  161, 42, 7, 1;
  1666, 378, 56, 7, 1; <-- row 5 of T
  28119, 5348, 679, 70, 7, 1; ...
where row 5 of T = row 4 of T^7 with appended '1'.
ALTERNATE GENERATING METHOD.
Row 4: start with a '1' followed by 4 zeros;
take partial sums and append 2 zeros; then
take partial sums thrice more:
  (1), 0, 0, 0, 0;
  1, 1, 1, 1, (1), 0, 0;
  1, 2, 3, 4, 5, 5, (5);
  1, 3, 6, 10, 15, 20, (25);
  1, 4, 10, 20, 35, 55, (80);
the final nonzero terms form row 4: [80, 25, 5, 1, 1].
Row 5: start with a '1' followed by 6 zeros;
take partial sums and append 4 zeros;
take partial sums and append 2 zeros; then
take partial sums thrice more:
  (1), 0, 0, 0, 0, 0, 0;
  1, 1, 1, 1, 1, 1, (1), 0, 0, 0, 0;
  1, 2, 3, 4, 5, 6, 7, 7, 7, 7, (7), 0, 0;
  1, 3, 6, 10, 15, 21, 28, 35, 42, 49, 56, 56, (56);
  1, 4, 10, 20, 35, 56, 84, 119, 161, 210, 266, 322, (378);
  1, 5, 15, 35, 70, 126, 210, 329, 490, 700, 966, 1288, (1666);
the final nonzero terms form row 5: [1666, 378, 56, 7, 1, 1].
Continuing in this way produces all the rows of this triangle.
		

Crossrefs

Cf. columns: A132616, A132617, A132618; A132619; variants: A132610, A101479.

Programs

  • Maple
    b:= proc(n) option remember;
          Matrix(n, (i,j)-> T(i-1,j-1))^(2*n-3)
        end:
    T:= proc(n,k) option remember;
         `if`(n=k, 1, `if`(k>n, 0, b(n)[n,k+1]))
        end:
    seq(seq(T(n,k), k=0..n), n=0..10);  # Alois P. Heinz, Apr 13 2020
  • Mathematica
    b[n_] := b[n] = MatrixPower[Table[T[i-1, j-1], {i, n}, {j, n}], 2n-3];
    T[n_, k_] := T[n, k] = If[n == k, 1, If[k > n, 0, b[n][[n, k + 1]]]];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 27 2020, after Alois P. Heinz *)
  • PARI
    {T(n, k)=local(A=vector(n+1), p); A[1]=1; for(j=1, n-k-1, p=(n-1)*(n-2)-(n-j-1)*(n-j-2); A=Vec((Polrev(A)+x*O(x^p))/(1-x))); A=Vec((Polrev(A)+x*O(x^p))/(1-x)); A[p+1]}

Formula

T(n+1,1) is divisible by 2n-1 for n>=1.

A304192 G.f. A(x) satisfies: [x^n] (1+x)^(n*(n+1)) / A(x) = 0 for n>0.

Original entry on oeis.org

1, 2, 7, 72, 1224, 29184, 892074, 33144288, 1445847756, 72291575784, 4070550314292, 254674699992768, 17518238545282080, 1313558965998605568, 106608039857256267192, 9309469431887521270848, 870250987085629018699728, 86703492688056304091302944, 9171254392641669833788501488, 1026466161170552167031522911104
Offset: 0

Views

Author

Paul D. Hanna, May 07 2018

Keywords

Comments

Note that: [x^n] (1+x)^(n*k) / G(x) = 0 for n>0 holds when G(x) = (1+x)/(1 - (k-1)*x) given some fixed k ; this sequence explores the case where k varies with n.

Examples

			G.f.: A(x) = 1 + 2*x + 7*x^2 + 72*x^3 + 1224*x^4 + 29184*x^5 + 892074*x^6 + 33144288*x^7 + 1445847756*x^8 + 72291575784*x^9 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^(n*(n+1)) / A(x) begins:
n=0: [1, -2, -3, -52, -955, -24246, -771113, -29428232, ...];
n=1: [1, 0, -6, -60, -1062, -26208, -820560, -30994704, ...];
n=2: [1, 4, 0, -80, -1337, -30840, -932010, -34438500, ...];
n=3: [1, 10, 39, 0, -1722, -39996, -1138680, -40521096, ...];
n=4: [1, 18, 147, 648, 0, -50832, -1503546, -50844384, ...];
n=5: [1, 28, 372, 3048, 15465, 0, -1898490, -67990260, ...];
n=6: [1, 40, 774, 9580, 83248, 483240, 0, -85539792, ...];
n=7: [1, 54, 1425, 24420, 303363, 2844270, 18685905, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] (1+x)^(n*(n+1)) / A(x) = 0 for n>0.
RELATED SEQUENCES.
The secondary diagonal in the above table that begins
[1, 4, 39, 648, 15465, 483240, 18685905, 861282832, 46085893011, ...]
yields A132612, column 1 of triangle A132610.
Related triangular matrix T = A132610 begins:
1;
1, 1;
2, 1, 1;
14, 4, 1, 1;
194, 39, 6, 1, 1;
4114, 648, 76, 8, 1, 1;
118042, 15465, 1510, 125, 10, 1, 1;
4274612, 483240, 41121, 2908, 186, 12, 1, 1;
186932958, 18685905, 1424178, 89670, 4970, 259, 14, 1, 1; ...
in which  row n+1 of T = row n of matrix power T^(2*n) with appended '1' for n>=0.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1,n, A=concat(A,0); m=#A; A[m] = Vec( (1+x +x*O(x^m))^(m*(m-1))/Ser(A) )[m] ); A[n+1]}
    for(n=0,30, print1(a(n),", "))

Formula

A132612(n+1) = [x^n] (1+x)^((n+1)*(n+2)) / A(x) for n>0.

A304193 G.f. A(x) satisfies: [x^n] (1+x)^((n+1)^2) / A(x) = 0 for n>0.

Original entry on oeis.org

1, 4, 16, 144, 2346, 55236, 1688084, 63040736, 2770165274, 139623836116, 7925496107656, 499719554537584, 34625595715906866, 2613946666882042164, 213475621178226876156, 18748792440158256161216, 1761875767691411063734514, 176383456081424163875684516, 18739798321516251204837796864, 2105891800817103192582808107856
Offset: 0

Views

Author

Paul D. Hanna, May 07 2018

Keywords

Comments

Note that: [x^n] (1+x)^((n+1)*k) / G(x) = 0 for n>0 holds when G(x) = (1+x)^(k+1)/(1 - (k-1)*x) given some fixed k ; this sequence explores the case where k varies with n.

Examples

			G.f.: A(x) = 1 + 4*x + 16*x^2 + 144*x^3 + 2346*x^4 + 55236*x^5 + 1688084*x^6 + 63040736*x^7 + 2770165274*x^8 + 139623836116*x^9 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^((n+1)^2) / A(x) begins:
n=0: [1, -3, -4, -80, -1530, -40222, -1316104, -51439572, ...];
n=1: [1, 0, -10, -100, -1785, -45056, -1441440, -55510080, ...];
n=2: [1, 5, 0, -140, -2380, -55080, -1685620, -63186200, ...];
n=3: [1, 12, 56, 0, -3150, -74484, -2125948, -76230384, ...];
n=4: [1, 21, 200, 1020, 0, -96492, -2901052, -98301840, ...];
n=5: [1, 32, 486, 4540, 26015, 0, -3718000, -135081440, ...];
n=6: [1, 45, 980, 13640, 132810, 855478, 0, -172046940, ...];
n=7: [1, 60, 1760, 33520, 462150, 4790156, 34461260, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] (1+x)^((n+1)^2) / A(x) = 0 for n>0.
RELATED SEQUENCES.
The secondary diagonal in the above table that begins
[1, 5, 56, 1020, 26015, 855478, 34461260, 1642995124, ...]
yields A132618, column 2 of triangle A132615.
Related triangular matrix T = A132615 begins:
1;
1, 1;
1, 1, 1;
6, 3, 1, 1;
80, 25, 5, 1, 1;
1666, 378, 56, 7, 1, 1;
47232, 8460, 1020, 99, 9, 1, 1;
1694704, 252087, 26015, 2134, 154, 11, 1, 1;
73552752, 9392890, 855478, 61919, 3848, 221, 13, 1, 1; ...
in which row n equals row (n-1) of T^(2*n-1) followed by '1' for n > 0.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1,n, A=concat(A,0); m=#A; A[m] = Vec( (1+x +x*O(x^m))^(m^2)/Ser(A) )[m] ); A[n+1]}
    for(n=0,30, print1(a(n),", "))

Formula

A132618(n+1) = [x^n] (1+x)^((n+2)^2) / A(x) for n>=0.

A304190 G.f. A(x) satisfies: [x^n] (1+x)^((n-1)^2) / A(x) = 0 for n>0.

Original entry on oeis.org

1, 0, 0, 4, 90, 2448, 79716, 3058740, 135637242, 6835557984, 386119895256, 24170805494868, 1661105052140226, 124342746871407984, 10070793262851698412, 877493877654988612836, 81848857574562663295026, 8137513480199793111630528, 859067817713438540813133744, 95973644392465888508242272804, 11312379843382901418721437545706
Offset: 0

Views

Author

Paul D. Hanna, May 08 2018

Keywords

Examples

			G.f.: A(x) = 1 + 4*x^3 + 90*x^4 + 2448*x^5 + 79716*x^6 + 3058740*x^7 + 135637242*x^8 + 6835557984*x^9 + 386119895256*x^10 + 24170805494868*x^11 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^((n-1)^2) / A(x) begins:
n=0: [1, 1, 0, -4, -94, -2538, -82148, -3137720, ...];
n=1: [1, 0, 0, -4, -90, -2448, -79700, -3058020, ...];
n=2: [1, 1, 0, -4, -94, -2538, -82148, -3137720, ...];
n=3: [1, 4, 6, 0, -105, -2832, -90048, -3391872, ...];
n=4: [1, 9, 36, 80, 0, -3276, -105224, -3871476, ...];
n=5: [1, 16, 120, 556, 1666, 0, -123900, -4673220, ...];
n=6: [1, 25, 300, 2296, 12460, 47232, 0, -5561820, ...];
n=7: [1, 36, 630, 7136, 58671, 368784, 1694704, 0,  ...];
n=8: [1, 49, 1176, 18420, 211590, 1895322, 13604628, 73552752, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] (1+x)^((n-1)^2) / A(x) = 0 for n>0.
RELATED SEQUENCES.
The secondary diagonal in the above table that begins
[1, 1, 6, 80, 1666, 47232, 1694704, 73552752, ...]
yields A132616, column 0 of triangle A132615.
Related triangular matrix T = A132615 begins:
1;
1, 1;
1, 1, 1;
6, 3, 1, 1;
80, 25, 5, 1, 1;
1666, 378, 56, 7, 1, 1;
47232, 8460, 1020, 99, 9, 1, 1;
1694704, 252087, 26015, 2134, 154, 11, 1, 1;
73552752, 9392890, 855478, 61919, 3848, 221, 13, 1, 1; ...
in which row n equals row (n-1) of T^(2*n-1) followed by '1' for n > 0.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( (1+x +x*O(x^m))^((m-2)^2)/Ser(A) )[m] ); A[n+1]}
    for(n=0, 30, print1(a(n), ", "))

Formula

A132616(n+1) = [x^n] (1+x)^(n^2) / A(x) for n>=0.

A304643 G.f. A(x) satisfies: [x^n] (1+x)^(n^3) / A(x) = 0 for n>0.

Original entry on oeis.org

1, 1, 21, 2075, 427745, 150754575, 80775206341, 61079788584715, 61918201760905701, 81032697606275994779, 132999148265782603510745, 267549402517056738883934727, 647439631215495429552890390761, 1855591663455916911410267165824087, 6216559993885861267930628826256971069, 24072412148295906199113974687972130690707, 106699538321376193436754733217464490904934733
Offset: 0

Views

Author

Paul D. Hanna, May 15 2018

Keywords

Examples

			G.f.: A(x) = 1 + x + 21*x^2 + 2075*x^3 + 427745*x^4 + 150754575*x^5 + 80775206341*x^6 + 61079788584715*x^7 + 61918201760905701*x^8 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^(n^3)/A(x) begins:
n=0: [1, -1, -20, -2034, -423216, -149819400, -80452969380, ...];
n=1: [1, 0, -21, -2054, -425250, -150242616, -80602788780, ...];
n=2: [1, 7, 0, -2166, -440034, -153263214, -81663489960, ...];
n=3: [1, 26, 304, 0, -470529, -161955486, -84652727166, ...];
n=4: [1, 63, 1932, 36334, 0, -174849912, -90924716676, ...];
n=5: [1, 124, 7605, 305466, 8541159, 0, -98844355155, ...];
n=6: [1, 215, 22984, 1626786, 85217850, 3329937702, 0, ...];
n=7: [1, 342, 58290, 6599344, 557724906, 37306986588, 1944420120804, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] (1+x)^(n^3) / A(x) = 0 for n>0.
RELATED SERIES.
1 - 1/A(x) = x + 20*x^2 + 2034*x^3 + 423216*x^4 + 149819400*x^5 + 80452969380*x^6 + 60910650903564*x^7 + 61792107766345152*x^8 + ...
The logarithmic derivative of the g.f. A(x) begins
A'(x)/A(x) = 1 + 41*x + 6163*x^2 + 1701881*x^3 + 751428751*x^4 + 483682989449*x^5 + 426965933360359*x^6 + 494840882952869729*x^7 + ...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( (1+x +x*O(x^m))^((m-1)^3)/Ser(A) )[m] ); A[n+1]}
    for(n=0, 20, print1(a(n), ", "))

A304644 G.f. A(x) satisfies: [x^n] (1+x)^(n^4) / A(x) = 0 for n>0.

Original entry on oeis.org

1, 1, 105, 73865, 149937065, 663916103529, 5451834603894529, 74704077908738108545, 1585534054417382287240065, 49309970434271232435701612225, 2152501158830776821954197582557961, 127436616988374904669593064111888541481, 9949767410829299590962659524265243208970825, 1000853528058644375639385529872204384996958065865, 127177120321862418629253989604625620834052796464647105
Offset: 0

Views

Author

Paul D. Hanna, May 15 2018

Keywords

Examples

			G.f.: A(x) = 1 + x + 105*x^2 + 73865*x^3 + 149937065*x^4 + 663916103529*x^5 + 5451834603894529*x^6 + 74704077908738108545*x^7 + 1585534054417382287240065*x^8 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^(n^4) / A(x) begins:
n=0: [1, -1, -104, -73656, -149778624, -663600972000, -5450470326008280, ...];
n=1: [1, 0, -105, -73760, -149852280, -663750750624, -5451133926980280, ...];
n=2: [1, 15, 0, -74880, -150968340, -666006324396, -5461105956428160, ...];
n=3: [1, 80, 3055, 0, -154503300, -675956601408, -5504713445922300, ...];
n=4: [1, 255, 32280, 2630600, 0, -696001081248, -5625102954138200, ...];
n=5: [1, 624, 194271, 40161344, 6040383876, 0, -5818032088967780, ...];
n=6: [1, 1295, 837760, 360910080, 116308352940, 29159359047060, 0, ...];
n=7: [1, 2400, 2878695, 2300795040, 1378317489120, 659313875405856, 255975781942704720, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] (1+x)^(n^4) / A(x) = 0 for n>0.
RELATED SERIES.
1 - 1/A(x) = x + 104*x^2 + 73656*x^3 + 149778624*x^4 + 663600972000*x^5 + 5450470326008280*x^6 + 74693014771268857320*x^7 + 1585383397658861643763200*x^8 + ...
The logarithmic derivative of the g.f. A(x) begins
A'(x)/A(x) = 1 + 209*x + 221281*x^2 + 599431169*x^3 + 3318792477121*x^4 + 32706914292746129*x^5 + 522889821925387405441*x^6 + 12683669785848215443184129*x^7 + ...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( (1+x +x*O(x^m))^((m-1)^4)/Ser(A) )[m] ); A[n+1]}
    for(n=0, 20, print1(a(n), ", "))
Showing 1-6 of 6 results.