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 1-10 of 20 results. Next

A181819 Prime shadow of n: a(1) = 1; for n>1, if n = Product prime(i)^e(i), then a(n) = Product prime(e(i)).

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 7, 2, 6, 2, 6, 4, 4, 2, 10, 3, 4, 5, 6, 2, 8, 2, 11, 4, 4, 4, 9, 2, 4, 4, 10, 2, 8, 2, 6, 6, 4, 2, 14, 3, 6, 4, 6, 2, 10, 4, 10, 4, 4, 2, 12, 2, 4, 6, 13, 4, 8, 2, 6, 4, 8, 2, 15, 2, 4, 6, 6, 4, 8, 2, 14, 7, 4, 2, 12, 4, 4, 4, 10, 2, 12, 4, 6, 4, 4, 4, 22, 2, 6, 6, 9, 2, 8, 2, 10, 8
Offset: 1

Views

Author

Matthew Vandermast, Dec 07 2010

Keywords

Comments

a(n) depends only on prime signature of n (cf. A025487). a(m) = a(n) iff m and n have the same prime signature, i.e., iff A046523(m) = A046523(n).
Because A046523 (the smallest representative of prime signature of n) and this sequence are functions of each other as A046523(n) = A181821(a(n)) and a(n) = a(A046523(n)), it implies that for all i, j: a(i) = a(j) <=> A046523(i) = A046523(j) <=> A101296(i) = A101296(j), i.e., that equivalence-class-wise this is equal to A101296, and furthermore, applying any function f on this sequence gives us a sequence b(n) = f(a(n)) whose equivalence class partitioning is equal to or coarser than that of A101296, i.e., b is then a sequence that depends only on the prime signature of n (the multiset of exponents of its prime factors), although not necessarily in a very intuitive way. - Antti Karttunen, Apr 28 2022

Examples

			20 = 2^2*5 has the exponents (2,1) in its prime factorization. Accordingly, a(20) = prime(2)*prime(1) = A000040(2)*A000040(1) = 3*2 = 6.
		

Crossrefs

Programs

Formula

From Antti Karttunen, Feb 07 2016: (Start)
a(1) = 1; for n > 1, a(n) = A000040(A067029(n)) * a(A028234(n)).
a(1) = 1; for n > 1, a(n) = A008578(A001511(n)) * a(A064989(n)).
Other identities. For all n >= 1:
a(A124859(n)) = A122111(a(n)) = A238745(n). - from Matthew Vandermast's formulas for the latter sequence.
(End)
a(n) = A246029(A156552(n)). - Antti Karttunen, Oct 15 2016
From Antti Karttunen, Apr 28 & Apr 30 2022: (Start)
A181821(a(n)) = A046523(n) and a(A046523(n)) = a(n). [See comments]
a(n) = A329900(A124859(n)) = A319626(A124859(n)).
a(n) = A246029(A156552(n)).
a(a(n)) = A328830(n).
a(A304660(n)) = n.
a(A108951(n)) = A122111(n).
a(A185633(n)) = A322312(n).
a(A025487(n)) = A181820(n).
a(A276076(n)) = A275735(n) and a(A276086(n)) = A328835(n).
As the sequence converts prime exponents to prime indices, it effects the following mappings:
A001221(a(n)) = A071625(n). [Number of distinct indices --> Number of distinct exponents]
A001222(a(n)) = A001221(n). [Number of indices (i.e., the number of prime factors with multiplicity) --> Number of exponents (i.e., the number of distinct prime factors)]
A056239(a(n)) = A001222(n). [Sum of indices --> Sum of exponents]
A066328(a(n)) = A136565(n). [Sum of distinct indices --> Sum of distinct exponents]
A003963(a(n)) = A005361(n). [Product of indices --> Product of exponents]
A290103(a(n)) = A072411(n). [LCM of indices --> LCM of exponents]
A156061(a(n)) = A290107(n). [Product of distinct indices --> Product of distinct exponents]
A257993(a(n)) = A134193(n). [Index of the least prime not dividing n --> The least number not among the exponents]
A055396(a(n)) = A051904(n). [Index of the least prime dividing n --> Minimal exponent]
A061395(a(n)) = A051903(n). [Index of the greatest prime dividing n --> Maximal exponent]
A008966(a(n)) = A351564(n). [All indices are distinct (i.e., n is squarefree) --> All exponents are distinct]
A007814(a(n)) = A056169(n). [Number of occurrences of index 1 (i.e., the 2-adic valuation of n) --> Number of occurrences of exponent 1]
A056169(a(n)) = A136567(n). [Number of unitary prime divisors --> Number of exponents occurring only once]
A064989(a(n)) = a(A003557(n)) = A295879(n). [Indices decremented after <--> Exponents decremented before]
Other mappings:
A007947(a(n)) = a(A328400(n)) = A329601(n).
A181821(A007947(a(n))) = A328400(n).
A064553(a(n)) = A000005(n) and A000005(a(n)) = A182860(n).
A051903(a(n)) = A351946(n).
A003557(a(n)) = A351944(n).
A258851(a(n)) = A353379(n).
A008480(a(n)) = A309004(n).
a(A325501(n)) = A325507(n) and a(A325502(n)) = A038754(n+1).
a(n!) = A325508(n).
(End)

Extensions

Name "Prime shadow" (coined by Gus Wiseman in A325755) prefixed to the definition by Antti Karttunen, Apr 27 2022

A071626 Number of distinct exponents in the prime factorization of n!.

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 8, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11
Offset: 1

Views

Author

Labos Elemer, May 29 2002

Keywords

Comments

Erdős proved that there exist two constants c1, c2 > 0 such that c1 (n / log(n))^(1/2) < a(n) < c2 (n / log(n))^(1/2). - Carlo Sanna, May 28 2019
R. Heyman and R. Miraj proved that the cardinality of the set { floor(n/p) : p <= n, p prime } is same as the number of distinct exponents in the prime factorization of n!. - Md Rahil Miraj, Apr 05 2024

Examples

			n=7: 7! = 5040 = 2*2*2*2*3*3*5*7; three different exponents arise: 4, 2 and 1; a(7)=3.
n=7: { floor(7/p) : p <= 7, p prime } = {3,2,1}. So, its cardinality is 3. - _Md Rahil Miraj_, Apr 05 2024
		

Crossrefs

Programs

  • Mathematica
    ffi[x_] := Flatten[FactorInteger[x]] lf[x_] := Length[FactorInteger[x]] ep[x_] := Table[Part[ffi[x], 2*w], {w, 1, lf[x]}] Table[Length[Union[ep[w! ]]], {w, 1, 100}]
    Table[Length[Union[Last/@If[n==1,{},FactorInteger[n!]]]],{n,30}] (* Gus Wiseman, May 15 2019 *)
  • PARI
    a(n) = #Set(factor(n!)[, 2]); \\ Michel Marcus, Sep 05 2017

Formula

a(n) = A071625(n!) = A323023(n!,3). - Gus Wiseman, May 15 2019

A115627 Irregular triangle read by rows: T(n,k) = multiplicity of prime(k) as a divisor of n!.

Original entry on oeis.org

1, 1, 1, 3, 1, 3, 1, 1, 4, 2, 1, 4, 2, 1, 1, 7, 2, 1, 1, 7, 4, 1, 1, 8, 4, 2, 1, 8, 4, 2, 1, 1, 10, 5, 2, 1, 1, 10, 5, 2, 1, 1, 1, 11, 5, 2, 2, 1, 1, 11, 6, 3, 2, 1, 1, 15, 6, 3, 2, 1, 1, 15, 6, 3, 2, 1, 1, 1, 16, 8, 3, 2, 1, 1, 1, 16, 8, 3, 2, 1, 1, 1, 1
Offset: 2

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Author

Keywords

Comments

The factorization of n! is n! = 2^T(n,1)*3^T(n,2)*...*p_(pi(n))^T(n,pi(n)) where p_k = k-th prime, pi(n) = A000720(n).
Nonzero terms of A085604; T(n,k) = A085604(n,k), k = 1..A000720(n). - Reinhard Zumkeller, Nov 01 2013
For n=2, 3, 4 and 5, all terms of the n-th row are odd. Are there other such rows? - Michel Marcus, Nov 11 2018
From Gus Wiseman, May 15 2019: (Start)
Differences between successive rows are A067255, so row n is the sum of the first n row-vectors of A067255 (padded with zeros on the right so that all n row-vectors have length A000720(n)). For example, the first 10 rows of A067255 are
{}
1
0 1
2 0
0 0 1
1 1 0
0 0 0 1
3 0 0 0
0 2 0 0
1 0 1 0
with column sums (8,4,2,1), which is row 10.
(End)
For all prime p > 7, 3*p > 2*nextprime(p), so for any n > 21 there will always be a prime p dividing n! with exponent 2 and there are no further rows with all entries odd. - Charlie Neder, Jun 03 2019

Examples

			From _Gus Wiseman_, May 09 2019: (Start)
Triangle begins:
   1
   1  1
   3  1
   3  1  1
   4  2  1
   4  2  1  1
   7  2  1  1
   7  4  1  1
   8  4  2  1
   8  4  2  1  1
  10  5  2  1  1
  10  5  2  1  1  1
  11  5  2  2  1  1
  11  6  3  2  1  1
  15  6  3  2  1  1
  15  6  3  2  1  1  1
  16  8  3  2  1  1  1
  16  8  3  2  1  1  1  1
  18  8  4  2  1  1  1  1
(End)
m such that 5^m||101!: floor(log(101)/log(5)) = 2 terms. floor(101/5) = 20. floor(20/5) = 4. So m = u_1 + u_2 = 20 + 4 = 24. - _David A. Corneth_, Jun 22 2014
		

Crossrefs

Row lengths are A000720.
Row-sums are A022559.
Row-products are A135291.
Row maxima are A011371.

Programs

  • Haskell
    a115627 n k = a115627_tabf !! (n-2) !! (k-1)
    a115627_row = map a100995 . a141809_row . a000142
    a115627_tabf = map a115627_row [2..]
    -- Reinhard Zumkeller, Nov 01 2013
    
  • Maple
    A115627 := proc(n,k) local d,p; p := ithprime(k) ; n-add(d,d=convert(n,base,p)) ; %/(p-1) ; end proc: # R. J. Mathar, Oct 29 2010
  • Mathematica
    Flatten[Table[Transpose[FactorInteger[n!]][[2]], {n, 2, 20}]] (* T. D. Noe, Apr 10 2012 *)
    T[n_, k_] := Module[{p, jm}, p = Prime[k]; jm = Floor[Log[p, n]]; Sum[Floor[n/p^j], {j, 1, jm}]]; Table[Table[T[n, k], {k, 1, PrimePi[n]}], {n, 2, 20}] // Flatten (* Jean-François Alcover, Feb 23 2015 *)
  • PARI
    a(n)=my(i=2);while(n-primepi(i)>1,n-=primepi(i);i++);p=prime(n-1);sum(j=1,log(i)\log(p),i\=p) \\ David A. Corneth, Jun 21 2014

Formula

T(n,k) = Sum_{i=1..inf} floor(n/(p_k)^i). (Although stated as an infinite sum, only finitely many terms are nonzero.)
T(n,k) = Sum_{i=1..floor(log(n)/log(p_k))} floor(u_i) where u_0 = n and u_(i+1) = floor((u_i)/p_k). - David A. Corneth, Jun 22 2014

A325617 Multinomial coefficient of the prime signature of n!.

Original entry on oeis.org

1, 1, 1, 2, 4, 20, 105, 840, 3960, 51480, 675675, 10810800, 139675536, 2793510720, 58663725120, 1799020903680, 26985313555200, 782574093100800, 25992639520848000, 857757104187984000, 30021498646579440000, 1563341744336692320000, 64179292662243158400000
Offset: 0

Views

Author

Gus Wiseman, May 12 2019

Keywords

Comments

Number of permutations of the multiset of prime factors of n!.

Examples

			The a(5) = 20 permutations of {2,2,2,3,5}:
  (22235)  (32225)  (52223)
  (22253)  (32252)  (52232)
  (22325)  (32522)  (52322)
  (22352)  (35222)  (53222)
  (22523)
  (22532)
  (23225)
  (23252)
  (23522)
  (25223)
  (25232)
  (25322)
		

Crossrefs

Programs

  • Mathematica
    Table[Multinomial@@Last/@FactorInteger[n!],{n,0,15}]

Formula

a(n) = A318762(A181819(n!)).

A076934 Smallest integer of the form n/k!.

Original entry on oeis.org

1, 1, 3, 2, 5, 1, 7, 4, 9, 5, 11, 2, 13, 7, 15, 8, 17, 3, 19, 10, 21, 11, 23, 1, 25, 13, 27, 14, 29, 5, 31, 16, 33, 17, 35, 6, 37, 19, 39, 20, 41, 7, 43, 22, 45, 23, 47, 2, 49, 25, 51, 26, 53, 9, 55, 28, 57, 29, 59, 10, 61, 31, 63, 32, 65, 11, 67, 34, 69, 35, 71
Offset: 1

Views

Author

Amarnath Murthy, Oct 19 2002

Keywords

Comments

Equivalently, n divided by the largest factorial divisor of n.
Also, the smallest r such that n/r is a factorial number.
Positions of 1's are the factorial numbers A000142. Is every positive integer in this sequence? - Gus Wiseman, May 15 2019
Let m = A055874(n), the largest integer such that 1,2,...,m divides n. Then a(n*m!) = n since m+1 does not divide n, showing that every integer is part of the sequence. - Etienne Dupuis, Sep 19 2020

Crossrefs

Programs

  • Mathematica
    Table[n/Max@@Intersection[Divisors[n],Array[Factorial,n]],{n,100}] (* Gus Wiseman, May 15 2019 *)
    a[n_] := Module[{k=1}, While[Divisible[n, k!], k++]; n/(k-1)!]; Array[a, 100] (* Amiram Eldar, Dec 25 2023 *)
  • PARI
    first(n) = {my(res = [1..n]); for(i = 2, oo, k = i!; if(k <= n, for(j = 1, n\k, res[j*k] = j ) , return(res) ) ) } \\ David A. Corneth, Sep 19 2020

Formula

From Amiram Eldar, Dec 25 2023: (Start)
a(n) = n/A055881(n)!.
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = BesselI(2, 2) = 0.688948... (A229020). (End)

Extensions

More terms from David A. Corneth, Sep 19 2020

A325509 Number of factorizations of n! into factorial numbers > 1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 3, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Gus Wiseman, May 08 2019

Keywords

Examples

			n = 10:
  (6*120*5040)
  (720*5040)
  (3628800)
n = 16:
  (2*2*2*2*1307674368000)
  (2*120*87178291200)
  (20922789888000)
n = 24:
  (2*2*6*25852016738884976640000)
  (24*25852016738884976640000)
  (620448401733239439360000)
		

Crossrefs

Programs

  • Mathematica
    facs[n_,u_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d,u],Min@@#>=d&]],{d,Intersection[u,Rest[Divisors[n]]]}]];
    Table[Length[facs[n!,Rest[Array[#!&,n]]]],{n,15}]

Formula

a(n) = 1 + A034876(n).

Extensions

More terms from Alois P. Heinz, May 08 2019

A307035 a(n) is the unique integer k such that A108951(k) = n!.

Original entry on oeis.org

1, 1, 2, 3, 12, 20, 60, 84, 672, 1512, 5040, 7920, 47520, 56160, 157248, 393120, 6289920, 8225280, 37013760, 41368320, 275788800, 579156480, 1820206080, 2203407360, 26440888320, 73446912000, 173601792000, 585906048000, 3281073868800, 4137006182400, 20685030912000
Offset: 0

Views

Author

Allan C. Wechsler, Mar 20 2019

Keywords

Comments

For all n, n! = A108951(k) for some unique k. This sequence gives that k for each n. In some sense this sequence tells how to factor factorials into primorials.
Represent n! as a product of primorials p#. Then replace each primorial with its base prime to calculate a(n).

Examples

			Represent 7! as a product of primorials:
7! = 2^4 * 3^2 * 5 * 7 = (2#)^2 * 3# * 7#
Replace primorials by primes:
2^2 * 3 * 7 = 84.
So a(7) = 84.
		

Crossrefs

Programs

  • Maple
    f:= proc(n) option remember; `if`(n<2, 0, f(n-1)+add(
          i[2]*x^numtheory[pi](i[1]), i=ifactors(n)[2]))
        end:
    a:= proc(n) local d, p, r; p, r:= f(n), 1;
          do d:= degree(p); if d<1 then break fi;
             p, r:= p-add(x^i, i=1..d), ithprime(d)*r
          od: r
        end:
    seq(a(n), n=0..35);  # Alois P. Heinz, Mar 21 2019
  • Mathematica
    q[n_] := Apply[Times, Table[Prime[i], {i, 1, PrimePi[n]}]]; Flatten[{1, 1, Table[val = 1; fak = n!; Do[If[PrimeQ[k], Do[If[Divisible[fak, q[k]], val = val*k; fak = fak/q[k]], {j, 1, n}]], {k, n, 2, -1}]; val, {n, 2, 30}]}] (* Vaclav Kotesovec, Mar 21 2019 *)
  • PARI
    g(n) = my(f=factor(n)); prod(k=1, #f~, my(p=f[k, 1]); (p/if(p>2, precprime(p-1), 1))^f[k, 2]); \\ A319626/A319627
    a(n) = prod(k=1, n, g(k)); \\ Daniel Suteu, Mar 21 2019
    
  • PARI
    A307035(n) = { my(m=1, pp=1); n=n!; while(1, forprime(p=2, ,if(n%p, if(2==p, return(m), break), n /= p; pp = p)); m *= pp); }; \\ Antti Karttunen, Dec 29 2019

Formula

a(0) = 1, a(n) = a(n-1) * (A319626(n) / A319627(n)), for n > 0. - Daniel Suteu, Mar 21 2019
a(n) = n! / Product_{k=1..n} A064989(k). - Vaclav Kotesovec, Mar 21 2019
a(n) = A122111(A325508(n)) = A319626(A000142(n)) = A329900(A000142(n)). - Antti Karttunen, Nov 19 & Dec 29 2019

Extensions

a(12)-a(13) from Michel Marcus, Mar 21 2019
a(14)-a(15) from Vaclav Kotesovec, Mar 21 2019
a(0), a(16)-a(30) from Alois P. Heinz, Mar 21 2019

A325619 Heinz numbers of integer partitions whose reciprocal factorial sum is 1.

Original entry on oeis.org

2, 9, 375, 15625
Offset: 1

Views

Author

Gus Wiseman, May 13 2019

Keywords

Comments

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).
The reciprocal factorial sum of an integer partition (y_1,...,y_k) is 1/y_1! + ... + 1/y_k!.

Examples

			The sequence of terms together with their prime indices begins:
      1: {}
      2: {1}
      9: {2,2}
    375: {2,3,3,3}
  15625: {3,3,3,3,3,3}
		

Crossrefs

Reciprocal factorial sum: A002966, A051908, A316855, A325618, A325624.

Programs

  • Mathematica
    Select[Range[100000],Total[Cases[FactorInteger[#],{p_,k_}:>k/PrimePi[p]!]]==1&]

Formula

Contains prime(n)^(n!) for all n > 0, including 191581231380566414401 for n = 4.

A325620 Number of integer partitions of n whose reciprocal factorial sum is an integer.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 5, 5, 5, 6, 7, 7, 8, 9, 10, 10, 11, 12, 14, 14, 15, 16, 18, 19, 20, 22, 24, 25, 26, 28, 31, 33, 34, 36, 39, 41, 43, 45, 49, 52, 54, 57, 61, 65, 68, 71, 76, 80, 84, 88, 93, 98, 103, 107, 113
Offset: 1

Views

Author

Gus Wiseman, May 13 2019

Keywords

Comments

The reciprocal factorial sum of an integer partition (y_1,...,y_k) is 1/y_1! + ... + 1/y_k!.

Examples

			The initial terms count the following partitions:
  1: (1)
  2: (1,1)
  3: (1,1,1)
  4: (2,2)
  4: (1,1,1,1)
  5: (2,2,1)
  5: (1,1,1,1,1)
  6: (2,2,1,1)
  6: (1,1,1,1,1,1)
  7: (2,2,1,1,1)
  7: (1,1,1,1,1,1,1)
  8: (2,2,2,2)
  8: (2,2,1,1,1,1)
  8: (1,1,1,1,1,1,1,1)
  9: (2,2,2,2,1)
  9: (2,2,1,1,1,1,1)
  9: (1,1,1,1,1,1,1,1,1)
		

Crossrefs

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n],IntegerQ[Total[1/(#!)]]&]],{n,30}]

A325622 Number of integer partitions of n whose reciprocal factorial sum is the reciprocal of an integer.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 2, 3, 2, 3, 3, 2, 2, 3, 3, 3, 5, 4, 4, 3, 3, 4, 6, 3, 4, 5, 5, 5, 6, 3, 7, 6, 5, 6, 6, 6, 5, 6, 8, 5, 7, 5, 4, 8, 7, 7, 7, 7, 9, 9, 9, 10, 12, 6, 12, 8, 10, 7, 14, 10, 8, 11, 11, 12, 11, 10, 10, 12, 14, 11, 10, 9, 10, 12, 10, 15, 14, 11, 10
Offset: 1

Views

Author

Gus Wiseman, May 13 2019

Keywords

Comments

The reciprocal factorial sum of an integer partition (y_1,...,y_k) is 1/y_1! + ... + 1/y_k!.

Examples

			The initial terms count the following partitions:
   1: (1)
   2: (2)
   3: (3)
   4: (4)
   4: (2,2)
   5: (5)
   6: (6)
   6: (3,3)
   7: (7)
   8: (8)
   8: (4,4)
   9: (9)
   9: (5,4)
   9: (3,3,3)
  10: (10)
  10: (5,5)
  11: (11)
  11: (4,4,3)
  11: (3,3,3,2)
  12: (12)
  12: (6,6)
  12: (4,4,4)
		

Crossrefs

Reciprocal factorial sum: A002966, A316854, A316857, A325618, A325620, A325623.

Programs

  • Maple
    f:= proc(n) nops(select(proc(t) local i; (1/add(1/i!,i=t))::integer end proc, combinat:-partition(n))) end proc:
    map(f, [$1..70]); # Robert Israel, May 09 2024
  • Mathematica
    Table[Length[Select[IntegerPartitions[n],IntegerQ[1/Total[1/(#!)]]&]],{n,30}]
  • PARI
    a(n) = my(c=0); forpart(v=n, if(numerator(sum(i=1, #v, 1/v[i]!))==1, c++)); c; \\ Jinyuan Wang, Feb 25 2025

Extensions

a(61)-a(70) from Robert Israel, May 09 2024
a(71)-a(80) from Jinyuan Wang, Feb 25 2025
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