A008607 -id:A008607 - cp's OEIS frontend

A195040 Square array read by antidiagonals with T(n,k) = kn^2/4+(k-4)((-1)^n-1)/8, n>=0, k>=0.

0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 3, 2, 1, 0, 1, 4, 5, 3, 1, 0, 0, 7, 8, 7, 4, 1, 0, 1, 9, 13, 12, 9, 5, 1, 0, 0, 13, 18, 19, 16, 11, 6, 1, 0, 1, 16, 25, 27, 25, 20, 13, 7, 1, 0, 0, 21, 32, 37, 36, 31, 24, 15, 8, 1, 0, 1, 25, 41, 48, 49, 45, 37, 28, 17, 9, 1, 0

Offset: 0

Omar E. Pol, Sep 27 2011

Also, if k >= 2 and m = 2*k, then column k lists the numbers of the form k*n^2 and the centered m-gonal numbers interleaved.
For k >= 3, this is also a table of concentric polygonal numbers. Column k lists the concentric k-gonal numbers.
It appears that the first differences of column k are the numbers that are congruent to {1, k-1} mod k, if k >= 3.

			Array begins:
  0,   0,   0,   0,   0,   0,   0,   0,   0,   0, ...
  1,   1,   1,   1,   1,   1,   1,   1,   1,   1, ...
  0,   1,   2,   3,   4,   5,   6,   7,   8,   9, ...
  1,   3,   5,   7,   9,  11,  13,  15,  17,  19, ...
  0,   4,   8,  12,  16,  20,  24,  28,  32,  36, ...
  1,   7,  13,  19,  25,  31,  37,  43,  49,  55, ...
  0,   9,  18,  27,  36,  45,  54,  63,  72,  81, ...
  1,  13,  25,  37,  49,  61,  73,  85,  97, 109, ...
  0,  16,  32,  48,  64,  80,  96, 112, 128, 144, ...
  1,  21,  41,  61,  81, 101, 121, 141, 161, 181, ...
  0,  25,  50,  75, 100, 125, 150, 175, 200, 225, ...
  ...

Muniru A Asiru, Rows n=0..100, flattened

Rows n: A000004 (n=0), A000012 (n=1), A001477 (n=2), A005408 (n=3), A008586 (n=4), A016921 (n=5), A008591 (n=6), A017533 (n=7), A008598 (n=8), A215145 (n=9), A008607 (n=10).

Columns k: A000035 (k=0), A004652 (k=1), A000982 (k=2), A077043 (k=3), A000290 (k=4), A032527 (k=5), A032528 (k=6), A195041 (k=7), A077221 (k=8), A195042 (k=9), A195142 (k=10), A195043 (k=11), A195143 (k=12), A195045 (k=13), A195145 (k=14), A195046 (k=15), A195146 (k=16), A195047 (k=17), A195147 (k=18), A195048 (k=19), A195148 (k=20), A195049 (k=21), A195149 (k=22), A195058 (k=23), A195158 (k=24).

nmax:=13;; T:=List([0..nmax],n->List([0..nmax],k->k*n^2/4+(k-4)*((-1)^n-1)/8));; b:=List([2..nmax],n->OrderedPartitions(n,2));;
a:=Flat(List([1..Length(b)],i->List([1..Length(b[i])],j->T[b[i][j][2]][b[i][j][1]]))); # Muniru A Asiru, Jul 19 2018

A195040 := proc(n,k)
        k*n^2/4+((-1)^n-1)*(k-4)/8 ;
end proc:
for d from 0 to 12 do
        for k from 0 to d do
                printf("%d,",A195040(d-k,k)) ;
        end do:
end do; # R. J. Mathar, Sep 28 2011

t[n_, k_] := k*n^2/4+(k-4)*((-1)^n-1)/8; Flatten[ Table[ t[n-k, k], {n, 0, 11}, {k, 0, n}]] (* Jean-François Alcover, Dec 14 2011 *)

A016850 a(n) = (5*n)^2.

Original entry on oeis.org

0, 25, 100, 225, 400, 625, 900, 1225, 1600, 2025, 2500, 3025, 3600, 4225, 4900, 5625, 6400, 7225, 8100, 9025, 10000, 11025, 12100, 13225, 14400, 15625, 16900, 18225, 19600, 21025, 22500, 24025, 25600, 27225, 28900, 30625, 32400, 34225, 36100, 38025, 40000, 42025

Offset: 0

N. J. A. Sloane

If we define C(n) = (5*n)^2 (n > 0), the sequence is the first "square-sequence" such that for every n there exists p such that C(n) = C(p) + C(p+n). We observe in fact that p = 3*n because 25 = 3^2 + 4^2. The sequence without 0 is linked with the first nontrivial solution (trivial: n^2 = 0^2 + n^2) of the equation X^2 = 2*Y^2 + 2*n^2 where X = 2*k and Y = 2*p + n which is equivalent to k^2 = p^2 + (p+n)^2 for n given. The second such "square-sequence" is (29*n)^2 (n > 0) because 29^2 = 20^2 + 21^2 and with this relation we obtain (29*n)^2 = (20*n)^2 + (20*n+n)^2. - Richard Choulet, Dec 23 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..800
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A033429, A053742 (first differences), A008587, A008607.

Similar sequences listed in A244630.

[(5*n)^2: n in [0..50]]; // Vincenzo Librandi, Apr 26 2011

(5Range[0, 31])^2 (* Alonso del Arte, Oct 08 2017 *)

a(n)=(5*n)^2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 25*n^2 = 25*A000290(n) = 5*A033429(n). - Omar E. Pol, Jul 03 2014
From Amiram Eldar, Jan 25 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/150.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/300.
Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/5)/(Pi/5).
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/5)/(Pi/5) = 5*sqrt((5-sqrt(5))/2)/(2*Pi). (End)
a(n) = Sum_{i=0..n-1} A053742(i). - John Elias, Jun 30 2021
G.f.: 25*x*(1 + x)/(1 - x)^3. - Stefano Spezia, Jul 08 2023
From Elmo R. Oliveira, Nov 30 2024: (Start)
E.g.f.: 25*x*(1 + x)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
a(n) = n*A008607(n) = A000290(A008587(n)) = A008587(n)^2. (End)

A262221 a(n) = 25n(n + 1)/2 + 1.

Original entry on oeis.org

1, 26, 76, 151, 251, 376, 526, 701, 901, 1126, 1376, 1651, 1951, 2276, 2626, 3001, 3401, 3826, 4276, 4751, 5251, 5776, 6326, 6901, 7501, 8126, 8776, 9451, 10151, 10876, 11626, 12401, 13201, 14026, 14876, 15751, 16651, 17576, 18526, 19501, 20501, 21526, 22576, 23651

Offset: 0

Bruno Berselli, Sep 15 2015

Also centered 25-gonal (or icosipentagonal) numbers.
This is the case k=25 of the formula (k*n*(n+1) - (-1)^k + 1)/2. See table in Links section for similar sequences.
For k=2*n, the formula shown above gives A011379.
Primes in sequence: 151, 251, 701, 1951, 3001, 4751, 10151, 12401, ...

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 51 (23rd row of the table).

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Table of numbers of the form (k*n*(n+1)-(-1)^k+1)/2.
Eric Weisstein's World of Mathematics, Centered Polygonal Numbers.
Index entries for sequences related to centered polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A008607 (first differences), A011379, A034856, A054254, A080956, A123296, A186349, A255184.
Cf. centered polygonal numbers listed in A069190.
Similar sequences of the form (k*n*(n+1) - (-1)^k + 1)/2 with -1 <= k <= 26: A000004, A000124, A002378, A005448, A005891, A028896, A033996, A035008, A046092, A049598, A060544, A064200, A069099, A069125, A069126, A069128, A069130, A069132, A069174, A069178, A080956, A124080, A163756, A163758, A163761, A164136, A173307.

Magma
```
[25*n*(n+1)/2+1: n in [0..50]];
```

Table[25 n (n + 1)/2 + 1, {n, 0, 50}]
25*Accumulate[Range[0,50]]+1 (* or *) LinearRecurrence[{3,-3,1},{1,26,76},50] (* Harvey P. Dale, Jan 29 2023 *)

PARI
```
vector(50, n, n--; 25*n*(n+1)/2+1)
```
Sage
```
[25*n*(n+1)/2+1 for n in (0..50)]
```

G.f.: (1 + 23*x + x^2)/(1 - x)^3.
a(n) = a(-n-1) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = A123296(n) + 1.
a(n) = A000217(5*n+2) - 2.
a(n) = A034856(5*n+1).
a(n) = A186349(10*n+1).
a(n) = A054254(5*n+2) with n>0, a(0)=1.
a(n) = A000217(n+1) + 23*A000217(n) + A000217(n-1) with A000217(-1)=0.
Sum_{i>=0} 1/a(i) = 1.078209111... = 2*Pi*tan(Pi*sqrt(17)/10)/(5*sqrt(17)).
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=0} a(n)/n! = 77*e/2.
Sum_{n>=0} (-1)^(n+1) * a(n)/n! = 23/(2*e). (End)
E.g.f.: exp(x)*(2 + 50*x + 25*x^2)/2. - Elmo R. Oliveira, Dec 24 2024

A252994 Multiples of 26.

Original entry on oeis.org

0, 26, 52, 78, 104, 130, 156, 182, 208, 234, 260, 286, 312, 338, 364, 390, 416, 442, 468, 494, 520, 546, 572, 598, 624, 650, 676, 702, 728, 754, 780, 806, 832, 858, 884, 910, 936, 962, 988, 1014, 1040, 1066, 1092, 1118, 1144, 1170, 1196, 1222, 1248, 1274, 1300, 1326, 1352, 1378

Offset: 0

Alonso del Arte, Dec 25 2014

Richard A. Mollin, Codes: The Guide to Secrecy From Ancient to Modern Times. Chapman and Hall/CRC (2005): 12.
Anne L. Young, Mathematical Ciphers: From Caesar to RSA. American Mathematical Society (2006): 23.

Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A005843, A008595, A008607, A244633.

Mathematica
```
26 Range[0, 51]
```
PARI
```
a(n)=26*n \\ Georg Fischer, Feb 25 2019
```

a(n) = 26*n.
a(n) = 2*a(n-1) - a(n-2).
G.f.: 26*x/(1-x)^2. - Georg Fischer, Feb 25 2019
From Elmo R. Oliveira, Apr 08 2025: (Start)
E.g.f.: 26*x*exp(x).
a(n) = A005843(A008595(n)). (End)

A317325 Multiples of 25 and odd numbers interleaved.

Original entry on oeis.org

0, 1, 25, 3, 50, 5, 75, 7, 100, 9, 125, 11, 150, 13, 175, 15, 200, 17, 225, 19, 250, 21, 275, 23, 300, 25, 325, 27, 350, 29, 375, 31, 400, 33, 425, 35, 450, 37, 475, 39, 500, 41, 525, 43, 550, 45, 575, 47, 600, 49, 625, 51, 650, 53, 675, 55, 700, 57, 725, 59, 750, 61, 775, 63, 800, 65, 825, 67, 850, 69

Offset: 0

Omar E. Pol, Jul 25 2018

Partial sums give the generalized 29-gonal numbers (A303815).

a(n) is also the length of the n-th line segment of the rectangular spiral whose vertices are the generalized 29-gonal numbers.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A008607 and A005408 interleaved.
Column 25 of A195151.
Sequences whose partial sums give the generalized k-gonal numbers: A026741 (k=5), A001477 (k=6), zero together with A080512 (k=7), A022998 (k=8), A195140 (k=9), zero together with A165998 (k=10), A195159 (k=11), A195161 (k=12), A195312 (k=13), A195817 (k=14).
Cf. A303815.

Flat(List([0..40],n->[25*n,2*n+1])); # Muniru A Asiru, Jul 28 2018

&cat[[25*n, 2*n + 1]: n in [0..30]]; // Vincenzo Librandi, Jul 28 2018

seq(op([25*n,2*n+1]),n=0..40); # Muniru A Asiru, Jul 28 2018

With[{nn=30}, Riffle[25 Range[0, nn], 2 Range[0, nn] + 1]] (* Vincenzo Librandi, Jul 28 2018 *)

concat(0, Vec(x*(1 + 25*x + x^2) / ((1 - x)^2*(1 + x)^2) + O(x^60))) \\ Colin Barker, Jul 29 2018

a(2n) = 25*n, a(2n+1) = 2*n + 1.
G.f.: x*(1 + 25*x + x^2)/((1 - x)^2*(1 + x)^2). - Vincenzo Librandi, Jul 28 2018
a(n) = 2*a(n-2) - a(n-4) for n>3. - Colin Barker, Jul 29 2018
Multiplicative with a(2^e) = 25*2^(e-1), and a(p^e) = p^e for an odd prime p. - Amiram Eldar, Oct 14 2023
Dirichlet g.f.: zeta(s-1) * (1 + 23/2^s). - Amiram Eldar, Oct 26 2023

A305548 a(n) = 27*n.

Original entry on oeis.org

0, 27, 54, 81, 108, 135, 162, 189, 216, 243, 270, 297, 324, 351, 378, 405, 432, 459, 486, 513, 540, 567, 594, 621, 648, 675, 702, 729, 756, 783, 810, 837, 864, 891, 918, 945, 972, 999, 1026, 1053, 1080, 1107, 1134, 1161, 1188, 1215, 1242, 1269, 1296, 1323, 1350, 1377, 1404, 1431, 1458, 1485, 1512

Offset: 0

Eric Chen, Jun 05 2018

Eric Chen, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1).

For a(n) = k*n: A001489 (k=-1), A000004 (k=0), A001477 (k=1), A005843 (k=2), A008585 (k=3), A008591 (k=9), A008607 (k=25), A252994 (k=26), this sequence (k=27), A135628 (k=28), A195819 (k=29), A249674 (k=30), A135631 (k=31), A174312 (k=32), A044102 (k=36), A085959 (k=37), A169823 (k=60), A152691 (k=64).

Mathematica
```
Range[0,2000,27]
```
PARI
```
a(n)=27*n
```

a(n) = 27*n.
a(n) = A008585(A008591(n)) = A008591(A008585(n)).
G.f.: 27*x/(x-1)^2.
From Elmo R. Oliveira, Apr 10 2025: (Start)
E.g.f.: 27*x*exp(x).
a(n) = 2*a(n-1) - a(n-2). (End)

A351381 Table read by downward antidiagonals: T(n,k) = n*(k+1)^2.

Original entry on oeis.org

4, 9, 8, 16, 18, 12, 25, 32, 27, 16, 36, 50, 48, 36, 20, 49, 72, 75, 64, 45, 24, 64, 98, 108, 100, 80, 54, 28, 81, 128, 147, 144, 125, 96, 63, 32, 100, 162, 192, 196, 180, 150, 112, 72, 36, 121, 200, 243, 256, 245, 216, 175, 128, 81, 40, 144, 242, 300, 324, 320, 294, 252, 200, 144, 90, 44

Offset: 1

Bernard Schott, Mar 28 2022

When m and k are both positive integers and k | m, with m/k = n, then T(n,k) = S(m,k) = (m+k) + (m-k) + (m*k) + (m/k) = S(n*k,k) = n*(k+1)^2, problem proposed by Yakov Perelman.

All terms are nonsquarefree (A013929).

			Table begins:
  n \ k |   1      2      3      4      5      6      7      8      9     10
  ----------------------------------------------------------------------------
     1  |   4      9     16      25    36     49     64     81    100    121
     2  |   8     18     32      50    72     98    128    162    200    242
     3  |  12     27     48      75   108    147    192    243    300    363
     4  |  16     36     64     100   144    196    256    324    400    484
     5  |  20     45     80     125   180    245    320    405    500    605
     6  |  24     54     96     150   216    294    384    486    600    726
     7  |  28     63    112     175   252    343    448    567    700    847
     8  |  32     72    128     200   288    392    512    648    800    968
     9  |  36     81    144     225   324    441    576    729    900   1089
    10  |  40     90    160     250   360    490    640    810   1000   1210
  ............................................................................
T(3,4) = 75 = 3*(4+1)^2 corresponds to S(3*4,4) = S(12,4) = (12+4) + (12-4) + (12*4) + 12/4 = 75.
S(10,5) = (10+5) + (10-5) + (10*5) + (10/5) = T(10/5,5) = T(2,5) = 72.

I. Perelman, L'Algèbre Récréative, Chapitre IV, Les équations de Diophante, Deux nombres et quatre opérations, Editions en langues étrangères, Moscou, 1959, pp. 101-102.
Ya. I. Perelman, Algebra Can Be Fun, Chapter IV, Diophantine Equations, Two numbers and four operations, Mir Publishers Moscow, 1979, pp. 131-132.

Ya. I. Perelman, Algebra Can Be Fun, Chapter IV, Diophantine Equations, Two numbers and four operations, Mir Publishers Moscow, 1979, pp. 131-132.
Wikipedia, Yakov Perelman.

Cf. A013929.
Cf. A000290 \ {0,1} (row 1), A001105 \ {0,2} (row 2), A033428 \ {0,3} (row 3), A016742 \ {0,4} (row 4), A033429 \ {0,5} (row 5), A033581 \ {0,6} (row 6).
Cf. A008586 \ {0} (column 1), A008591 \ {0} (column 2), A008598 \ {0} (column 3), A008607 \ {0} (column 4), A044102 \ {0} (column 5).
Cf. A045991 \ {0} (diagonal).

T[n_, k_] := n*(k + 1)^2; Table[T[k, n - k + 1], {n, 1, 11}, {k, 1, n}] // Flatten (* Amiram Eldar, Mar 29 2022 *)

T(n,k) = n*(k+1)^2.
T(n,n) = (n+1)^3 - (n+1)^2 = A045991(n+1) for n >= 1.
G.f.: x*(1 + y)/((1 - x)^2*(1 - y)^3). - Stefano Spezia, Mar 31 2022

A044482 Numbers k such that the string 4,4 occurs in the base-5 representation of k but not of k+1.

Original entry on oeis.org

24, 49, 74, 99, 124, 149, 174, 199, 224, 249, 274, 299, 324, 349, 374, 399, 424, 449, 474, 499, 524, 549, 574, 624, 649, 674, 699, 724, 749, 774, 799, 824, 849, 874, 899, 924, 949, 974, 999, 1024, 1049, 1074, 1099, 1124, 1149

Offset: 1

Clark Kimberling

a(n)+1 is divisible by 25 because the string 4,4 must be at the lower end of the base-5 representation. - Joerg Arndt, Feb 13 2012

			49 is in the sequence because its base-5 representation, 144, includes 44 as a substring, but the base-5 representation of 50, 200, does not.
599 is not in the sequence because, although it is 4344 in base 5, 600 is 4400 in base 5.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A008607 (multiples of 25).

f[n_]:= Length[StringPosition[ToString[FromDigits[IntegerDigits[n,5]]],"44",1]]; Select[Table[n,{n,2000}],f[#]>0&&f[#+1] == 0&] (* Vincenzo Librandi, Feb 12 2012 *)
SequencePosition[Table[If[SequenceCount[IntegerDigits[n,5],{4,4}]>0,1,0],{n,1200}],{1,0}][[All,1]] (* Harvey P. Dale, Aug 20 2021 *)

A377165 a(n) = 76 + 275*n.

Original entry on oeis.org

76, 351, 626, 901, 1176, 1451, 1726, 2001, 2276, 2551, 2826, 3101, 3376, 3651, 3926, 4201, 4476, 4751, 5026, 5301, 5576, 5851, 6126, 6401, 6676, 6951, 7226, 7501, 7776, 8051, 8326, 8601, 8876, 9151, 9426, 9701, 9976, 10251, 10526, 10801, 11076, 11351, 11626, 11901

Offset: 0

Stefano Spezia, Oct 18 2024

Numbers k such k - 1 is a multiple of 25 and k + 1 is a multiple of 11. See Besso and Lugli, and Matmedia links.

Davide Besso and Aurelio Lugli, Temi di Matematica dati per l'Esame di Maturità in Ginnasi e Scuole Reali Superiori dell'Austria-Ungheria, Periodico di Matematica, ISSN 2612-6745, Series I, 1894. See p. 27 (In Italian).
Matmedia, Un esercizio di maturità sui numeri interi, Oct 08 2024 (In Italian).
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A008593, A008607, A377166.

Mathematica
```
LinearRecurrence[{2,-1},{76,351},44]
```

a(n) = 2*a(n-1) - a(n-2) for n > 1.
G.f.: (76 + 199*x)/(1 - x)^2.
E.g.f.: exp(x)*(76 + 275*x).

A119652 Number of different values of <= n standard American coins (pennies, nickels, dimes and quarters).

Original entry on oeis.org

4, 13, 27, 46, 69, 94, 119, 144, 169, 194, 219, 244, 269, 294, 319, 344, 369, 394, 419, 444, 469, 494, 519, 544, 569, 594, 619, 644, 669, 694, 719, 744, 769, 794, 819, 844, 869, 894, 919, 944, 969, 994, 1019, 1044, 1069, 1094, 1119, 1144, 1169, 1194, 1219

Offset: 1

Tanya Khovanova, Jul 28 2006

nonn

			If you have 1 coin you can have 4 different totals: 1, 5, 10 and 25. If you have 2 coins, you can have 10 totals: 2, 6, 10, 11, 15, 20, 26, 30, 35, 50. Notice that the same total appears twice: 10 is one dime and two nickels. Hence a(2) = 13.

Cf. A008607.

Join[{4, 13, 27, 46}, Range[69, 2000, 25]] (* Vladimir Joseph Stephan Orlovsky, Jun 15 2011 *)

Conjectures from Colin Barker, Oct 25 2019: (Start)
G.f.: x*(4 + 5*x + 5*x^2 + 5*x^3 + 4*x^4 + 2*x^5) / (1 - x)^2.
a(n) = 2*a(n-1) - a(n-2) for n>6.
a(n) = 25*n-56 for n>4.
(End)

cp's OEIS Frontend

A195040 Square array read by antidiagonals with T(n,k) = k*n^2/4+(k-4)*((-1)^n-1)/8, n>=0, k>=0.

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GAP

Maple

Mathematica

A016850 a(n) = (5*n)^2.

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Magma

Mathematica

PARI

Formula

A262221 a(n) = 25*n*(n + 1)/2 + 1.

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Magma

Mathematica

PARI

Sage

Formula

A252994 Multiples of 26.

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Mathematica

PARI

Formula

A317325 Multiples of 25 and odd numbers interleaved.

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GAP

Magma

Maple

Mathematica

PARI

Formula

A305548 a(n) = 27*n.

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Programs

Mathematica

PARI

Formula

A351381 Table read by downward antidiagonals: T(n,k) = n*(k+1)^2.

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Author

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Comments

Examples

A195040 Square array read by antidiagonals with T(n,k) = kn^2/4+(k-4)((-1)^n-1)/8, n>=0, k>=0.

A262221 a(n) = 25n(n + 1)/2 + 1.