A051846 -id:A051846 - cp's OEIS frontend

A051847 Bisection of A051846, divided by the term position.

1, 19, 1493, 293479, 109739369, 66987982331, 60693710471869, 76519827268721103, 128138108936443028945, 275176672984400058317539, 737345594135016860806925221, 2411620538399461719230688945719

Offset: 1

Antti Karttunen, Dec 13 1999

Cf. A051846, A221740.

a(n) = A051846(2n-1)/(2n-1).

a(n) = A221740(2n-1). - Alexander R. Povolotsky, Oct 13 2022

A328290 Table T(b,n) = #{ k > 0 | nk has only distinct and nonzero digits in base b }, b >= 2, 1 <= n <= A051846(b).

Original entry on oeis.org

1, 4, 1, 0, 0, 1, 0, 1, 15, 5, 9, 0, 2, 3, 2, 0, 4, 1, 1, 0, 2, 1, 2, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 64, 42, 21, 9, 0, 14, 8, 4, 7, 0, 6, 4, 3, 6, 0, 3, 2, 5, 2, 0, 4, 5, 3, 2, 0, 0, 2, 1, 2, 0, 2, 1, 2, 1, 0, 1, 2, 1, 2, 0, 1, 3, 1, 1, 0, 1, 1, 2, 1, 0, 0, 0, 0, 2

Offset: 2

M. F. Hasler, Oct 11 2019

The table could also be considered as an infinite square array with T(b,n) = 0 for n > A051846(b) = the largest pandigital number in base b.

Can anyone find a simple formula for the index of the last terms > 1 in each row b?

			The table reads:  (column n >= 2 corresponds to the base)
   B \ n = 1      2      3      4      5      6      7     8      9      10  ...
   2       1     (0 ...)
   3       4      1      0      0      1      0      1    (0 ...)
   4      15      5      9      0      2      3      2     0      4       1  ...
   5      64     42     21      9      0     14      8     4      7       0  ...
   6     325    130     65     65    161      0     48    23     32      66  ...
   7    1956    651   1140    319    386    221      0   156    362     128  ...
   8   13699   5871   4506   1957   2748   1944   6277     0   1470    1189  ...
   9  109600  73588  27400  56930  21973  18397  15641  8305      0   14826  ...
  10  986409 438404 572175 219202 109601 255752 140515 109601 432645     0   ...
  (...)
In base 2, 1 is the only number with distinct nonzero digits, so T(2,1) = 1, T(2,n) = 0 for n > 1.
In base 3, {1, 2, 12_3 = 5, 21_3 = 7} are the only numbers with distinct nonzero digits, so T(3,1) = 4, T(3,2) = T(3,7) = T(3,7) = 1, T(3,n) = 0 for n > 7.
In base 4, {1, 2, 3, 12_4 = 6, 13_4 = 7, 21_4 = 9, ..., 321_4 = 57} are the only numbers with distinct nonzero digits, so T(4,n) = 0 for n > 57.

Cf. A328287 (row 10), A328288, A328277.

Column 1 is A007526 (number of nonnull variations of n distinct objects).

T(B,n)={my(S,T,U); for(L=1,B-1,T=vectorv(L,k,B^(k-1)); forperm(L,p, U=vecextract(T,p); forvec(D=vector(L,i,[1,B-1]),D*U%n||S++,2)));S}

T(b,b) = 0, since any multiple of b has a trailing digit 0 in base b.

T(b,A051846(b)) = 1 and T(b,n) = 0 for n > A051846(b) = (b-1)(b-2)..21 in base b.

A023811 Largest metadrome (number with digits in strict ascending order) in base n.

Original entry on oeis.org

0, 1, 5, 27, 194, 1865, 22875, 342391, 6053444, 123456789, 2853116705, 73686780563, 2103299351334, 65751519677857, 2234152501943159, 81985529216486895, 3231407272993502984, 136146740744970718253, 6106233505124424657789, 290464265927977839335179

Offset: 1

Olivier Gérard

Also smallest zeroless pandigital number in base n. - Franklin T. Adams-Watters, Nov 15 2006

The smallest permutational number in A134640 in the n-positional system. - Artur Jasinski, Nov 07 2007

			a(5) = 1234[5] (in base 5) = 1*5^3 + 2*5^2 + 3*5 + 4 = 125 + 50 + 15 + 4 = 194.
a(10) = 123456789 (in base 10).

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Christian Perfect, Integer sequence reviews on Numberphile (or vice versa), 2013.
Chai Wah Wu, Pandigital and penholodigital numbers, arXiv:2403.20304 [math.GM], 2024. See p. 1.

Cf. A049363, A051846, A058128, A060073.

Cf. A062813, A134640.

a023811 n = foldl (\val dig -> val * n + dig) 0 [0 .. n - 1]
-- Reinhard Zumkeller, Aug 29 2014

[0] cat [(n^n-n^2+n-1)/(n-1)^2: n in [2..20]]; // Vincenzo Librandi, May 22 2012

0, seq((n^n-n^2+n-1)/(n-1)^2, n=2..100); # Robert Israel, Dec 13 2015

Table[Total[(#1 n^#2) & @@@ Transpose@ {Range[n - 1], Reverse@ (Range[n - 1] - 1)}], {n, 20}] (* Michael De Vlieger, Jul 24 2015 *)
Table[Sum[(b - k)*b^(k - 1), {k, b - 1}], {b, 30}] (* Clark Kimberling, Aug 22 2015 *)
Table[FromDigits[Range[0, n - 1], n], {n, 20}] (* L. Edson Jeffery, Dec 13 2015 *)

{for(i=1,18,cuo=0; for(j=1,i-1,cuo=cuo+j*i^(i-j-1)); print1(cuo,", "))} \\\ Douglas Latimer, May 16 2012

A023811(n)=if(n>1,(n^n-n^2)\(n-1)^2+1)  \\ M. F. Hasler, Jan 22 2013

def a(n): return (n**n - n**2 + n - 1)//((n - 1)**2) if n > 1 else 0
print([a(n) for n in range(1, 21)]) # Michael S. Branicky, Apr 24 2023

a(n) = Sum_{j=1...n-1} j*n^(n-1-j).
lim_{n->infinity} a(n)/a(n-1) - a(n-1)/a(n-2) = exp(1). - Conjectured by Gerald McGarvey, Sep 26 2004. Follows from the formula below and lim_{n->infinity} (1+1/n)^n = e. - Franklin T. Adams-Watters, Jan 25 2010
a(n) = (n^n-n^2+n-1)/(n-1)^2 = A058128(n)-1 = n*A060073(n)-1 (for n>=2). - Henry Bottomley, Feb 21 2001

Edited by M. F. Hasler, Jan 22 2013

A051845 Triangle T(n,k) read by rows, in which row n gives all permutations of digits 1..n interpreted in base n+1.

Original entry on oeis.org

1, 5, 7, 27, 30, 39, 45, 54, 57, 194, 198, 214, 222, 238, 242, 294, 298, 334, 346, 358, 366, 414, 422, 434, 446, 482, 486, 538, 542, 558, 566, 582, 586, 1865, 1870, 1895, 1905, 1930, 1935, 2045, 2050, 2105, 2120, 2140, 2150, 2255, 2265, 2285, 2300, 2355, 2360

Offset: 1

Antti Karttunen, Dec 13 1999

All terms in any odd row 2m+1 are divisible by 2m+1
The n-th row has n! elements.
Variant of permutational numbers with shifted digits 0->1->2->...->p+1 in p+1 positional system -- see A134750. - Artur Jasinski, Nov 08 2007
From Alexander R. Povolotsky, Oct 22 2022: (Start)
All terms in any even-indexed row n=2m are divisible by m, where m>0.
Row n starts with T(n,1) = ((n+1)^(n+1)-n^2-n-1)/n^2 = A023811(n+1).
Row n ends with T(n,n!) = ((n+1)^(n+1)*(n-1)+1)/n^2 = A051846(n).
(End)

			Triangle begins:
         k=1   k=2   k=3  ...
  n=1:     1;
  n=2      5,    7;
  n=3:    27,   30,  ...,   57;
  n=4:   194,  198,  214,  ..., 586;
  n=5:  1865, 1870, 1905, 1930, ..., 7465;
E.g., the permutations of digits 1, 2 and 3 in lexicographic order are 123, 132, 213, 231, 312, 321, which interpreted in base 4 give the third row of the table: 27, 30, 39, 45, 54, 57.

Samuel Harkness, Table of n, a(n) for n = 1..10000

Left edge = A023811, right edge = A051846.

Cf. A134640, A134750.

with(combinat,permute); compute_u_rows := proc(u) local a,n; a := []; for n from 1 to u do a := [op(a),op(map(list_in_base_b,permute(n),(n+1)))]; od; RETURN(a); end; list_in_base_b := proc(l,b) local k; add(l[nops(l)-k]*(b^k), k=0..(nops(l)-1)); end;

a = {}; b = {}; Do[AppendTo[b, n]; w = Permutations[b]; Do[j = FromDigits[1 + w[[m]], n + 2]; AppendTo[a, j], {m, 1, Length[w]}], {n, 0, 5}]; a (* Artur Jasinski, Nov 08 2007 *)

from itertools import permutations
def fd(d, b): return sum(di*b**i for i, di in enumerate(d[::-1]))
def row(n): return [fd(d, n+1) for d in permutations(range(1, n+1))]
print([an for r in range(1, 6) for an in row(r)]) # Michael S. Branicky, Oct 21 2022

A221740 a(n) = -4((n-1)(n+1)^(n+1)+1)/(((-1)^n-3)*n^3).

Original entry on oeis.org

1, 7, 19, 293, 1493, 38127, 293479, 10593529, 109739369, 5135610071, 66987982331, 3856048810781, 60693710471869, 4149140360751583, 76519827268721103, 6058888636862818097, 128138108936443028945, 11533996620790579909159

Offset: 1

Alexander R. Povolotsky, Jan 23 2013

nonn

Per exhaustive program, written for bases from 2 to 10, the number of permutations pairs, which have the same ratio, equal to A221740(n)/A221741(n) = (n^2*(n+1)^n - (n+1)^n + 1) / (-n^2 + n*(n+1)^n + (n+1)^n - n - 1), is: {2,2,3,3,5,3,7,5,7,...} for n >= 1 where n = r-1 and r is the base radix. Judging by above sequence it appears that the number of such permutations pairs is related to phi, which is the Euler totient function - according to A039649, A039650, A214288 (see bullet 1 of the analysis in the answer section of the Mathematics StackExchange link). - Alexander R. Povolotsky, Jan 26 2013

G. C. Greubel, Table of n, a(n) for n = 1..385
Mathematics StackExchange, Permutations (with no duplicates) of decimal base digits 1,2,...,8,9,0.
NMBRTHRY, Number of specific permutations pairs relates to Euler Phi totient function?

Cf. A221741, A051846.

Table[-4*((n - 1)*(n + 1)^(n + 1) + 1)/(((-1)^n - 3)*n^3), {n,1,50}] (* G. C. Greubel, Feb 19 2017 *)

makelist(-4*((n-1)*(n+1)^(n+1)+1)/(((-1)^n-3)*n^3),n,1,20); /* Martin Ettl, Jan 25 2013 */

for(n=1,25, print1(-4*((n - 1)*(n + 1)^(n + 1) + 1)/(((-1)^n - 3)*n^3), ", ")) \\ G. C. Greubel, Feb 19 2017

a(n) = -4*A051846(n)/((-3 + (-1)^n)*n).
From Alexander R. Povolotsky, Oct 12 2022: (Start)
floor(a(n+1)/A221741(n+1)) = n.
Limit_{n->oo} (a(n)/A221741(n) - floor(a(n)/A221741(n))) = 0. (End)

A370371 Largest m such that any two consecutive digits of the base-n expansion of m^2 differ by 1 after arranging the digits in decreasing order.

Original entry on oeis.org

1, 1, 15, 2, 195, 867, 3213, 18858, 99066, 528905, 2950717, 294699, 105011842, 659854601, 4285181505, 1578809181, 198009443151, 1404390324525, 10225782424031, 3635290739033, 583655347579584, 4564790605900107, 36485812146621733, 297764406866494254, 2479167155959358950

Offset: 2

Jianing Song, Feb 16 2024

By definition, a(n) <= sqrt(Sum_{i=0..n-1} i*n^i) = sqrt(A062813(n)). If n is odd and n-1 has an even number of 2s as prime factors, then there are no pandigital squares in base n, so a(n) <= sqrt(Sum_{i=1..n-1} i*n^(i-1)) = sqrt(A051846(n-1)); see A258103.

If n is odd and n-1 has an even 2-adic valuation, then a(n) <= sqrt(Sum_{i=2..n-1} i*n^(i-2)); see A258103. - Chai Wah Wu, Feb 25 2024

			Base 4: 15^2 = 225 = 3201_4;
Base 6: 195^2 = 38025 = 452013_6;
Base 7: 867^2 = 751689 = 6250341_7;
Base 8: 3213^2 = 10323369 = 47302651_8;
Base 9: 18858^2 = 355624164 = 823146570_9;
Base 10: 99066^2 = 9814072356;
Base 11: 528905^2 = 279740499025 = A8701245369_11;
Base 12: 2950717^2 = 8706730814089 = B8750A649321_12;
Base 13: 294699^2 = 86847500601 = 8260975314_13.

Michael S. Branicky, Table of n, a(n) for n = 2..28

Cf. A215014, A370362, A370370, A258103 (number of pandigital squares in base n).
The actual squares are given by A370611.
Cf. A062813, A051846.

isconsecutive(m,n)=my(v=vecsort(digits(m,n))); for(i=2, #v, if(v[i]!=1+v[i-1], return(0))); 1 \\ isconsecutive(k,n) == 1 if and only if any two consecutive digits of the base-n expansion of m differ by 1 after arranging the digits in decreasing order
a(n) = forstep(m=sqrtint(if(n%2==1 && valuation(n-1, 2)%2==0, n^(n-1) - (n^(n-1)-1)/(n-1)^2, n^n - (n^n-n)/(n-1)^2)), 0, -1, if(isconsecutive(m^2,n), return(m)))

from math import isqrt
from sympy import multiplicity
from sympy.ntheory import digits
def a(n):
    ub = isqrt(sum(i*n**i for i in range(n)))
    if n%2 == 1 and multiplicity(2, n-1)%2 == 0:
        ub = isqrt(sum(i*n**(i-2) for i in range(2, n)))
    return(next(i for i in range(ub, -1, -1) if len(d:=sorted(digits(i*i, n)[1:])) == d[-1]-d[0]+1 == len(set(d))))
print([a(n) for n in range(2, 13)]) # Michael S. Branicky, Feb 23 2024

a(17)-a(20) and a(22)-a(26) from Michael S. Branicky, Feb 23 2024

a(21) from Chai Wah Wu, Feb 25 2024

A189001 a(n) = Sum_{i=0..n} (i+1)*n^i.

Original entry on oeis.org

1, 3, 17, 142, 1593, 22461, 380713, 7526268, 169826513, 4303999495, 120987654321, 3734729768298, 125562274081225, 4566262891748481, 178581127445062553, 7473240118999870456, 333189190735802745633, 15766036084935301064139

Offset: 0

Bruno Berselli, Apr 15 2011

nonn

			a(4) = 1593  because  1593 = 1+2*4+3*4^2+4*4^3+5*4^4.

Bruno Berselli, Table of n, a(n) for n = 0..300

Cf. A031973, A051846, A062806.

Cf. A189122: Sum_{i=0..n} (i+1)^2*n^i.

[&+[(k+1)*n^k: k in [0..n]]: n in [0..17]];

Join[{1, 3}, Table[(((n^2 - 2) n^(n + 1) + 1) / (n - 1)^2), {n, 2, 20}]] (* Vincenzo Librandi, Aug 19 2013 *)

a(n) = ((n^2-2)*n^(n+1)+1)/(n-1)^2 for n > 1; a(0)=1, a(1)=3.

A370611 Largest square such that any two consecutive digits of its base-n expansion differ by 1 after arranging the digits in decreasing order.

Original entry on oeis.org

1, 1, 225, 4, 38025, 751689, 10323369, 355624164, 9814072356, 279740499025, 8706730814089, 86847500601, 11027486960232964, 435408094460869201, 18362780530794065025, 2492638430009890761, 39207739576969100808801, 1972312183619434816475625, 104566626183621314286288961, 13215338757299095309775089

Offset: 2

Jianing Song, Feb 23 2024

By definition, a(n) <= Sum_{i=0..n-1} i*n^i = A062813(n). If n is odd and n-1 has an even number of 2s as prime factors, then there are no pandigital squares in base n, so a(n) <= Sum_{i=1..n-1} i*n^(i-1) = A051846(n-1); see A258103.

If n is odd and n-1 has an even 2-adic valuation, then a(n) <= Sum_{i=2..n-1} i*n^(i-2); see A258103. - Chai Wah Wu, Feb 25 2024

			See the Example section of A370371.

Cf. A215014, A370370, A370610, A258103 (number of pandigital squares in base n).
The square roots are given by A370371.
Cf. A062813, A051846.

isconsecutive(m, n)=my(v=vecsort(digits(m, n))); for(i=2, #v, if(v[i]!=1+v[i-1], return(0))); 1 \\ isconsecutive(k, n) == 1 if and only if any two consecutive digits of the base-n expansion of m differ by 1 after arranging the digits in decreasing order
a(n) = forstep(m=sqrtint(if(n%2==1 && valuation(n-1, 2)%2==0, n^(n-1) - (n^(n-1)-1)/(n-1)^2, n^n - (n^n-n)/(n-1)^2)), 0, -1, if(isconsecutive(m^2, n), return(m^2)))

a(17)-a(20) from Michael S. Branicky, Feb 23 2024

a(21) from Chai Wah Wu, Feb 25 2024

A363365 Array read by ascending antidiagonals: A(1, k) = k; for n > 1, A(n, k) = (k + 1)*A(n-1, k) + k + 1 - n, with k > 0.

Original entry on oeis.org

1, 2, 2, 3, 7, 3, 4, 21, 14, 4, 5, 62, 57, 23, 5, 6, 184, 228, 117, 34, 6, 7, 549, 911, 586, 207, 47, 7, 8, 1643, 3642, 2930, 1244, 333, 62, 8, 9, 4924, 14565, 14649, 7465, 2334, 501, 79, 9, 10, 14766, 58256, 73243, 44790, 16340, 4012, 717, 98, 10

Offset: 1

Stefano Spezia, May 29 2023

			The array begins:
  1,   2,    3,     4,     5, ...
  2,   7,   14,    23,    34, ...
  3,  21,   57,   117,   207, ...
  4,  62,  228,   586,  1244, ...
  5, 184,  911,  2930,  7465, ...
  6, 549, 3642, 14649, 44790, ...
  ...

Cf. A000027 (n=1 or k=1), A008865, A051846 (diagonal), A064017 (k=9), A353094 (k=2), A353095 (k=3), A353096 (k=4), A353097 (k=5), A353098 (k=6), A353099 (k=7), A353100 (k=8), A363366 (antidiagonal sums).

A[n_,k_]:=((k-1)*(k+1)^(n+1)+k*n-k^2+1)/k^2; Table[A[n-k+1,k],{n,10},{k,n}]//Flatten (* or *)
A[n_,k_]:=SeriesCoefficient[x*(k-(k+1)*x)/((1-x)^2*(1-(k+1)*x)),{x,0,n}]; Table[A[n-k+1,k],{n,10},{k,n}]//Flatten (* or *)
A[n_,k_]:=n!SeriesCoefficient[Exp[x]((k^2-1)(Exp[k x]-1)+k x)/k^2,{x,0,n}]; Table[A[n-k+1,k],{n,10},{k,n}]//Flatten

A(n, k) = ((k - 1)*(k + 1)^(n+1) + k*n - k^2 + 1)/k^2.
O.g.f. of k-th column: x*(k - (k + 1)*x)/((1 - x)^2*(1 - (k + 1)*x)).
E.g.f. of k-th column: exp(x)*((k^2 - 1)*(exp(k*x) - 1) + k*x)/k^2.
A(2, n) = A008865(n+1).

A370671 a(n) = Sum_{i=2..n-1} i*n^(i-2).

Original entry on oeis.org

0, 0, 0, 2, 14, 117, 1244, 16340, 256794, 4708235, 98765432, 2334368214, 61405650470, 1779714835745, 56358445438164, 1936265501684072, 71737338064426034, 2851241711464855575, 121019325106640638448, 5463472083532379956970, 261417839335180055401662, 13215375398730198560266829

Offset: 0

Chai Wah Wu, Feb 25 2024

nonn

Digits 2..(n-1) in strict descending order (n-1)..2 interpreted in base n.

Upper bound of A370371(n) when n is odd and n-1 has an even 2-adic valuation.

Cf. A062813, A051846, A370371.

def A370671(n): return (n-2)*(n**(n-1)-1)//(n-1)**2 if n > 1 else 0

a(n) = (n - 2)*(n^(n - 1) - 1)/(n - 1)^2 for n > 1.

cp's OEIS Frontend

A051847 Bisection of A051846, divided by the term position.

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A328290 Table T(b,n) = #{ k > 0 | nk has only distinct and nonzero digits in base b }, b >= 2, 1 <= n <= A051846(b).

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A023811 Largest metadrome (number with digits in strict ascending order) in base n.

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A051845 Triangle T(n,k) read by rows, in which row n gives all permutations of digits 1..n interpreted in base n+1.

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A221740 a(n) = -4*((n-1)*(n+1)^(n+1)+1)/(((-1)^n-3)*n^3).

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A370371 Largest m such that any two consecutive digits of the base-n expansion of m^2 differ by 1 after arranging the digits in decreasing order.

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A189001 a(n) = Sum_{i=0..n} (i+1)*n^i.

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A221740 a(n) = -4((n-1)(n+1)^(n+1)+1)/(((-1)^n-3)*n^3).