A000351 -id:A000351 - cp's OEIS frontend

A097863 Sum of 5-infinitary divisors of n.

1, 3, 4, 7, 6, 12, 8, 15, 13, 18, 12, 28, 14, 24, 24, 31, 18, 39, 20, 42, 32, 36, 24, 60, 31, 42, 40, 56, 30, 72, 32, 33, 48, 54, 48, 91, 38, 60, 56, 90, 42, 96, 44, 84, 78, 72, 48, 124, 57, 93, 72, 98, 54, 120, 72, 120, 80, 90, 60, 168, 62, 96, 104, 99, 84, 144, 68, 126, 96, 144, 72, 195, 74, 114, 124

Offset: 1

Yasutoshi Kohmoto

If n=Product p_i^r_i and d=Product p_i^s_i, each s_i has a digit a<=b in its 5-ary expansion everywhere that the corresponding r_i has a digit b, then d is a 5-infinitary-divisor of n.

			a(32) = a(2^10) = 2^10 + 2^0 = 32 + 1 = 33, in 5-ary expansion. This is the first term which is different from sigma(n).

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A049417, A049418, A074847, A097464.

Cf. A000351, A031235, A027748, A124010.

following Bower and Harris, cf. A049418:
a097863 1 = 1
a097863 n = product $ zipWith f (a027748_row n) (a124010_row n) where
   f p e = product $ zipWith div
           (map (subtract 1 . (p ^)) $
                zipWith (*) a000351_list $ map (+ 1) $ a031235_row e)
           (map (subtract 1 . (p ^)) a000351_list)
-- Reinhard Zumkeller, Sep 18 2015

A097863 := proc(n) option remember; local ifa, a, p, e, d, k ; ifa := ifactors(n)[2] ; a := 1 ; if nops(ifa) = 1 then p := op(1, op(1, ifa)) ; e := op(2, op(1, ifa)) ; d := convert(e, base, 5) ; for k from 0 to nops(d)-1 do a := a*(p^((1+op(k+1, d))*5^k)-1)/(p^(5^k)-1) ; end do: else for d in ifa do a := a*procname( op(1, d)^op(2, d)) ; end do: return a; end if; end proc:

f[p_, e_] := Module[{d = IntegerDigits[e, 5]}, m = Length[d]; Product[(p^((d[[j]] + 1)*5^(m - j)) - 1)/(p^(5^(m - j)) - 1), {j, 1, m}]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 09 2020 *)

Denote by P_5={p^5^k} the two-parameter set when k=0,1,... and p runs prime values. Then every n has a unique representation of the form n=prod q_i prod (r_j)^2 prod (s_k)^3 prod (t_m)^4, where q_i, r_j, s_k, t_m are distinct elements of P_5. Using this representation, we have a(n)=prod (q_i+1)prod ((r_j)^2+r_j+1)prod ((s_k)^3+(s_k)^2+s_k+1) prod ((t_m)^4+(t_m)^3+(t_m)^2+t_m+1). - Vladimir Shevelev, May 08 2013

A128963 a(n) = (n^3 - n)*5^n.

Original entry on oeis.org

0, 150, 3000, 37500, 375000, 3281250, 26250000, 196875000, 1406250000, 9667968750, 64453125000, 418945312500, 2666015625000, 16662597656250, 102539062500000, 622558593750000, 3735351562500000, 22178649902343750, 130462646484375000, 761032104492187500

Offset: 1

Mohammad K. Azarian, Apr 28 2007

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (20,-150,500,-625).

Cf. A000351, A007531, A036289, A081143, A128796.

Cf. A128960, A128961, A128962, A128964, A128965, A128967, A128969.

[(n^3-n)*5^n: n in [1..25]]; // Vincenzo Librandi, Feb 12 2013

Table[(n^3-n)5^n,{n,20}] (* or *) LinearRecurrence[{20,-150,500,-625},{0,150,3000,37500},20] (* Harvey P. Dale, Jul 22 2012 *)
CoefficientList[Series[150 x/(1 - 5 x)^4, {x, 0, 30}], x] (* Vincenzo Librandi, Feb 12 2013 *)

a(1)=0, a(2)=150, a(3)=3000, a(4)=37500, a(n)=20*a(n-1)-150*a(n-2)+ 500*a(n-3)- 625*a(n-4). - Harvey P. Dale, Jul 22 2012
G.f.: 150*x^2/(1 - 5*x)^4. - Vincenzo Librandi, Feb 12 2013
a(n) = 150*A081143(n+1). - Bruno Berselli, Feb 12 2013
From Amiram Eldar, Oct 02 2022: (Start)
a(n) = A007531(n+1)*A000351(n).
Sum_{n>=2} 1/a(n) = (8/5)*log(5/4) - 7/20.
Sum_{n>=2} (-1)^n/a(n) = (18/5)*log(6/5) - 13/20. (End)

Offset corrected by Mohammad K. Azarian, Nov 20 2008

A175552 Numbers k such that the digit sum of 167^k is divisible by k.

Original entry on oeis.org

1, 2, 5, 7, 22, 490, 724, 778, 868, 994, 1109, 1390, 1415, 1462, 1642, 1739, 1829, 2146, 2362, 3136, 4954, 6437, 6628, 7103, 11200, 12424, 12863, 14242, 14249, 15059, 15203, 16222, 17140, 18353, 19192, 21233, 22853, 24106, 24574, 24833, 26896, 27652, 28253, 30323, 31306, 31594, 32386, 33790, 34985, 36184, 36310, 40673, 42196, 43931, 45911, 45983

Offset: 1

N. J. A. Sloane, Dec 03 2010

From Donovan Johnson, Dec 03 2010: (Start)
To generate the additional terms I used PFGW.exe to get the decimal expansion for each number of the form 167^n (n <= 50000). Then I wrote a program in powerbasic to read the pfgw.out file and get the digit sums.
The digit sum is 10 times the n value for terms a(5) to a(56). (End)
I believe that this sequence is finite. - N. J. A. Sloane, Dec 05 2010

Donovan Johnson, Table of n, a(n) for n = 1..129

Sum of digits of k^n mod n: (k=2) A000079, A001370, A175434, A175169; (k=3) A000244, A004166, A175435, A067862; (k=5) A000351, A066001, A175456; (k=6) A000400, A066002, A175457, A067864; (k=7) A000420, A066003, A175512, A067863; (k=8) A062933; (k=13) A001022, A175527, A175528, A175525; (k=21) A175589; (k=167) A175558, A175559, A175560, A175552.

Select[Range[10000], Mod[Total[IntegerDigits[167^#]], #] == 0 &]

a(25)-a(56) from Donovan Johnson, Dec 03 2010

A191610 Possible number of trailing zeros in k!.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 62, 63, 64, 65, 66, 68, 69, 70, 71, 72, 74, 75, 76, 77, 78, 80, 81, 82, 83, 84, 86, 87, 88, 89, 90, 93, 94, 95, 96, 97, 99, 100, 101, 102, 103, 105, 106, 107, 108, 109, 111, 112, 113, 114, 115, 117, 118, 119, 120, 121, 124, 125, 126, 127, 128, 130, 131, 132, 133, 134, 136

Offset: 1

José María Grau Ribas, Jun 09 2011

nonn

Equivalently, possible values of 10-adic valuation of k!. - Joerg Arndt, Sep 21 2020

			3 is a term because 15! = 1307674368000 has 3 trailing 0's.
5 is not a term because 24! has 4 trailing 0's, but 25! has 6 trailing 0's.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A027868, A000351, A055457 (first differences).
Complement of A000966.
Cf. A380662, A381239.

a191610 1 = 0
a191610 n = sum $ takeWhile (> 0) $ map ((n - 1) `div`) a000351_list
-- Reinhard Zumkeller, Oct 31 2012

zOF[n_Integer?Positive]:=Module[{maxpow=0},While[5^maxpow<=n,maxpow++];Plus@@Table[ Quotient[n,5^i],{i,maxpow-1}]]; Attributes[zOF]={Listable}; zOF[Range[1000]]//Union (* Harvey P. Dale, Dec 06 2023 *)
Table[Sum[Floor[(n - 1)/5^k], {k, 0, Floor[Log[5, n]]}], {n, 1, 200}] (* Clark Kimberling, Feb 17 2025 *)

# requires Python 3.2 and higher
from itertools import accumulate
from sympy import multiplicity
A191610 = [0]+list(accumulate(multiplicity(5,n) for n in range(5,10**3,5)))
# Chai Wah Wu, Sep 05 2014

a(n) ~ 5*n/4. - Vaclav Kotesovec, Sep 21 2020
G.f.: 1/(1-x) * Sum_{k>=0} x^(5^k)/(1-x^5^k). - Joerg Arndt, Sep 21 2020
a(n) = Sum_{k>=0} floor((n-1)/5^k). - Clark Kimberling, Feb 17 2025

A212699 Main transitions in systems of n particles with spin 2.

Original entry on oeis.org

4, 40, 300, 2000, 12500, 75000, 437500, 2500000, 14062500, 78125000, 429687500, 2343750000, 12695312500, 68359375000, 366210937500, 1953125000000, 10375976562500, 54931640625000, 289916992187500, 1525878906250000, 8010864257812500, 41961669921875000, 219345092773437500

Offset: 1

Stanislav Sykora, May 25 2012

Please, refer to the general explanation in A212697.

This particular sequence is obtained for base b=5, corresponding to spin S=(b-1)/2=2.

Stanislav Sykora, Table of n, a(n) for n = 1..100
Stanislav Sýkora, Magnetic Resonance on OEIS, Stan's NMR Blog (Dec 31, 2014), Retrieved Nov 12, 2019.
Index entries for linear recurrences with constant coefficients, signature (10,-25).

Cf. A001787, A212697, A212698, A212700, A212701, A212702, A212703, A212704 (b = 2, 3, 4, 6, 7, 8, 9, 10).

Cf. A000351, A008586, A053464.

Join[{4},Table[4n*5^(n-1),{n,20}]] (* or *) Join[{4},LinearRecurrence[{10,-25},{4,40},20]] (* Harvey P. Dale, Aug 19 2014 *)

mtrans(n, b) = n*(b-1)*b^(n-1);
for (n=1, 100, write("b212699.txt", n, " ", mtrans(n, 5)))

a(n) = n*(b-1)*b^(n-1) where b=5.
a(n) = 10*a(n-1) - 25*a(n-2), a(0)=a(1)=4, a(2)=40. - Harvey P. Dale, Aug 19 2014
From Elmo R. Oliveira, May 13 2025: (Start)
G.f.: 4*x/(5*x-1)^2.
E.g.f.: 4*x*exp(5*x).
a(n) = 4*A053464(n) = A008586(n)*A000351(n-1). (End)

A319075 Square array T(n,k) read by antidiagonal upwards in which row n lists the n-th powers of primes, hence column k lists the powers of the k-th prime, n >= 0, k >= 1.

Original entry on oeis.org

1, 2, 1, 4, 3, 1, 8, 9, 5, 1, 16, 27, 25, 7, 1, 32, 81, 125, 49, 11, 1, 64, 243, 625, 343, 121, 13, 1, 128, 729, 3125, 2401, 1331, 169, 17, 1, 256, 2187, 15625, 16807, 14641, 2197, 289, 19, 1, 512, 6561, 78125, 117649, 161051, 28561, 4913, 361, 23, 1, 1024, 19683, 390625, 823543, 1771561, 371293

Offset: 0

Omar E. Pol, Sep 09 2018

If n = p - 1 where p is prime, then row n lists the numbers with p divisors.

The partial sums of column k give the column k of A319076.

			The corner of the square array is as follows:
         A000079 A000244 A000351  A000420    A001020    A001022     A001026
A000012        1,      1,      1,       1,         1,         1,          1, ...
A000040        2,      3,      5,       7,        11,        13,         17, ...
A001248        4,      9,     25,      49,       121,       169,        289, ...
A030078        8,     27,    125,     343,      1331,      2197,       4913, ...
A030514       16,     81,    625,    2401,     14641,     28561,      83521, ...
A050997       32,    243,   3125,   16807,    161051,    371293,    1419857, ...
A030516       64,    729,  15625,  117649,   1771561,   4826809,   24137569, ...
A092759      128,   2187,  78125,  823543,  19487171,  62748517,  410338673, ...
A179645      256,   6561, 390625, 5764801, 214358881, 815730721, 6975757441, ...
...

Rows 0-13: A000012, A000040, A001248, A030078, A030514, A050997, A030516, A092759, A179645, A179665, A030629, A079395, A030631, A138031.
Other rows n: A030635 (n=16), A030637 (n=18), A137486 (n=22), A137492 (n=28), A139571 (n=30), A139572 (n=36), A139573 (n=40), A139574 (n=42), A139575 (n=46), A173533 (n=52), A183062 (n=58), A183085 (n=60), A261700 (n=100).
Columns 1-15: A000079, A000244, A000351, A000420, A001020, A001022, A001026, A001029, A009967, A009973, A009975, A009981, A009985, A009987, A009991.
Main diagonal gives A093360.
Second diagonal gives A062457.
Third diagonal gives A197987.
Removing the 1's we have A182944/ A182945.
Cf. A006093, A319074, A319076.

PARI
```
T(n, k) = prime(k)^n;
```

T(n,k) = A000040(k)^n, n >= 0, k >= 1.

A013611 Triangle of coefficients in expansion of (1+4x)^n.

Original entry on oeis.org

1, 1, 4, 1, 8, 16, 1, 12, 48, 64, 1, 16, 96, 256, 256, 1, 20, 160, 640, 1280, 1024, 1, 24, 240, 1280, 3840, 6144, 4096, 1, 28, 336, 2240, 8960, 21504, 28672, 16384, 1, 32, 448, 3584, 17920, 57344, 114688, 131072, 65536, 1, 36, 576, 5376, 32256, 129024, 344064, 589824, 589824, 262144

Offset: 0

N. J. A. Sloane

T(n,k) equals the number of n-length words on {0,1,2,3,4} having n-k zeros. - Milan Janjic, Jul 24 2015

			Triangle begins
  1;
  1,    4;
  1,    8,   16;
  1,   12,   48,   64;
  1,   16,   96,  256,  256;
  1,   20,  160,  640, 1280, 1024;
  1,   24,  240, 1280, 3840, 6144, 4096;

Harvey P. Dale, Table of n, a(n) for n = 0..1034 (rows 0..44 flattened, missing terms added by Sean A. Irvine, Apr 21 2019)
J. Goldman, J. Haglund, Generalized rook polynomials, J. Combin. Theory A91 (2000), 509-530, 1-rook coefficients for k rooks on the 4xn board, all heights 4.

Cf. A000351 (5^n).

T:= n-> (p-> seq(coeff(p, x, k), k=0..n))((1+4*x)^n):
seq(T(n), n=0..10);  # Alois P. Heinz, Jul 24 2015

Flatten[Table[CoefficientList[Series[(1+4x)^n,{x,0,10}],x],{n,0,15}]] (* Harvey P. Dale, Oct 10 2011 *)

G.f.: 1 / (1 - x(1+4y)).

T(n,k) = 4^k*C(n,k) = Sum_{i=n-k..n} C(i,n-k)*C(n,i)*3^(n-i). Row sums are 5^n = A000351. - Mircea Merca, Apr 28 2012

A038243 Triangle whose (i,j)-th entry is 5^(i-j)*binomial(i,j).

Original entry on oeis.org

1, 5, 1, 25, 10, 1, 125, 75, 15, 1, 625, 500, 150, 20, 1, 3125, 3125, 1250, 250, 25, 1, 15625, 18750, 9375, 2500, 375, 30, 1, 78125, 109375, 65625, 21875, 4375, 525, 35, 1, 390625, 625000, 437500, 175000, 43750, 7000, 700, 40, 1, 1953125, 3515625, 2812500, 1312500, 393750, 78750, 10500, 900, 45, 1

Offset: 0

N. J. A. Sloane

Mirror image of A013612. - Zerinvary Lajos, Nov 25 2007
T(i,j) is the number of i-permutations of 6 objects a,b,c,d,e,f, with repetition allowed, containing j a's. - Zerinvary Lajos, Dec 21 2007
Triangle of coefficients in expansion of (5+x)^n - N-E. Fahssi, Apr 13 2008
Also the convolution triangle of A000351. - Peter Luschny, Oct 09 2022

			Triangle begins as:
       1;
       5,      1;
      25,     10,      1;
     125,     75,     15,      1;
     625,    500,    150,     20,     1;
    3125,   3125,   1250,    250,    25,    1;
   15625,  18750,   9375,   2500,   375,   30,   1;
   78125, 109375,  65625,  21875,  4375,  525,  35,  1;
  390625, 625000, 437500, 175000, 43750, 7000, 700, 40, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
B. N. Cyvin, J. Brunvoll, and S. J. Cyvin, Isomer enumeration of unbranched catacondensed polygonal systems with pentagons and heptagons, Match, No. 34 (Oct 1996), 109-121.

Cf. A000351, A000400 (row sums), A036071, A050982, A053464, A081135, A081143.

Sequences of the form q^(n-k)*binomial(n, k): A007318 (q=1), A038207 (q=2), A027465 (q=3), A038231 (q=4), this sequence (q=5), A038255 (q=6), A027466 (q=7), A038279 (q=8), A038291 (q=9), A038303 (q=10), A038315 (q=11), A038327 (q=12), A133371 (q=13), A147716 (q=14), A027467 (q=15).

[5^(n-k)*Binomial(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, May 12 2021

for i from 0 to 8 do seq(binomial(i, j)*5^(i-j), j = 0 .. i) od; # Zerinvary Lajos, Dec 21 2007
# Uses function PMatrix from A357368. Adds column 1, 0, 0, ... to the left.
PMatrix(10, n -> 5^(n-1)); # Peter Luschny, Oct 09 2022

With[{q=5}, Table[q^(n-k)*Binomial[n,k], {n,0,12}, {k,0,n}]//Flatten] (* G. C. Greubel, May 12 2021 *)

flatten([[5^(n-k)*binomial(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 12 2021

See A038207 and A027465 and replace 2 and 3 in analogous formulas with 5. - Tom Copeland, Oct 26 2012

A055842 Expansion of (1-x)^2/(1-5*x).

Original entry on oeis.org

1, 3, 16, 80, 400, 2000, 10000, 50000, 250000, 1250000, 6250000, 31250000, 156250000, 781250000, 3906250000, 19531250000, 97656250000, 488281250000, 2441406250000, 12207031250000, 61035156250000, 305175781250000, 1525878906250000, 7629394531250000

Offset: 0

Barry E. Williams, May 30 2000

First differences of A005054.
For n>=2, a(n) is equal to the number of functions f:{1,2,...,n}->{1,2,3,4,5} such that for fixed, different x_1, x_2 in {1,2,...,n} and fixed y_1, y_2 in {1,2,3,4,5} we have f(x_1)<>y_1 and f(x_2)<> y_2. - Milan Janjic, Apr 19 2007
a(n) is the number of generalized compositions of n when there are 4 *i-1 different types of i, (i=1,2,...). - Milan Janjic, Aug 26 2010

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (5).

Cf. A000351, A005054.

Concatenation([1,3], List([2..30], n-> 16*5^(n-2) )); # G. C. Greubel, Jan 21 2020

[1,3] cat [16*5^(n-2): n in [2..30]]; // G. C. Greubel, Jan 21 2020

R:=PowerSeriesRing(Rationals(), 25); Coefficients(R!( (1-x)^2/(1-5*x))); // Marius A. Burtea, Jan 21 2020

seq( `if`(n<2, 2*n+1, 16*5^(n-2)), n=0..30); # G. C. Greubel, Jan 21 2020

Join[{1,3},16 5^(Range[2,30]-2)] (* Harvey P. Dale, Apr 03 2013 *)

Vec((1-x)^2/(1-5*x) + O(x^30)) \\ Altug Alkan, Mar 13 2016

[1,3]+[16*5^(n-2) for n in (2..30)] # G. C. Greubel, Jan 21 2020

a(n) = 16*5^(n-2), a(0)=1, a(1)=3.
a(n) = 5*a(n-1) + (-1)^n*binomial(2,2-n).
G.f.: (1-x)^2/(1-5*x).
a(n) = Sum_{k=0..n} A201780(n,k)*3^k. - Philippe Deléham, Dec 05 2011
E.g.f.: (9 - 5*x + 16*exp(x))/25. - G. C. Greubel, Jan 21 2020

A056546 a(n) = 5na(n-1) + 1 with a(0)=1.

Original entry on oeis.org

1, 6, 61, 916, 18321, 458026, 13740781, 480927336, 19237093441, 865669204846, 43283460242301, 2380590313326556, 142835418799593361, 9284302221973568466, 649901155538149792621, 48742586665361234446576

Offset: 0

Henry Bottomley, Jun 20 2000

			a(2) = 5*2*a(1) + 1 = 10*6 + 1 = 61.

Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.

Cf. A000522, A010844, A010845, A056545, A056547 for analogs. A056546/(A000142*A000351) is an increasingly good approximation to 5th root of e.

m = 16; CoefficientList[E^x/(1-5x) + O[x]^m, x] Range[0, m-1]! (* Jean-François Alcover, Jun 03 2019 *)

a(n) = floor(e^(1/5)*5^n*n!).
From Philippe Deléham, Mar 14 2004: (Start)
a(n) = n!*Sum_{k=0..n} 5^(n-k)/k!.
E.g.f.: exp(x)/(1 - 5*x). (End)
a(n) = Sum_{k=0..n} P(n, k)*5^k. - Ross La Haye, Aug 29 2005
a(n) = hypergeometric_U(1, n+2 , 1/5)/5. - Peter Luschny, Nov 26 2014
From Peter Bala, Mar 01 2017: (Start)
a(n) = Integral_{x >= 0} (5*x + 1)^n*exp(-x) dx.
The e.g.f. y = exp(x)/(1 - 5*x) satisfies the differential equation (1 - 5*x)*y' = (6 - 5*x)*y.
a(n) = (5*n + 1)*a(n-1) - 5*(n - 1)*a(n-2).
The sequence b(n) := 5^n*n! also satisfies the same recurrence with b(0) = 1, b(1) = 5. This leads to the continued fraction representation a(n) = 5^n*n!*( 1 + 1/(5 - 5/(11 - 10/(16 - ... - (5*n - 5)/(5*n + 1) )))) for n >= 2. Taking the limit gives the continued fraction representation exp(1/5) = 1 + 1/(5 - 5/(11 - 10/(16 - ... - (5*n - 5)/((5*n + 1) - ... )))). Cf. A010844. (End)

More terms from James Sellers, Jul 04 2000

cp's OEIS Frontend

A097863 Sum of 5-infinitary divisors of n.

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A128963 a(n) = (n^3 - n)*5^n.

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A175552 Numbers k such that the digit sum of 167^k is divisible by k.

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A191610 Possible number of trailing zeros in k!.

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A212699 Main transitions in systems of n particles with spin 2.

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A319075 Square array T(n,k) read by antidiagonal upwards in which row n lists the n-th powers of primes, hence column k lists the powers of the k-th prime, n >= 0, k >= 1.

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A013611 Triangle of coefficients in expansion of (1+4x)^n.

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A056546 a(n) = 5na(n-1) + 1 with a(0)=1.