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%I #114 Feb 29 2020 17:59:54
%S 15,399,85,318,57,906,85,1670,1111,18193,185,7205205,4119,63791,4369,
%T 1548502,489,258099,451,408166,13315,1012985,679,25841526,26533,
%U 2884373,985,49101338,1057,5362755,1285,2447558,179503,3091422,1387,5830693854,82311,149338,2005
%N a(n) is the first composite number having the same base-n digits as its prime factors (with multiplicity), excluding zero digits (or 0 if no such composite number exists).
%C For an alternate program that only checks a single base at a time, use the code from "#the actual function (alternate)" instead of "#the actual function".
%C The computation of a(n) is exceedingly inefficient, requiring the checking of all natural values less than a(n). A more efficient way to compute a(n) is very desirable. - _Ely Golden_, Dec 25 2016
%C There is a lower bound on a(n), if not 0, of n^2 + n + 1. As well, a(n) must have 3 or more nonzero digits in base n (if n is odd, this lower bound is n^3 + n^2 + n + 1, and a(n) must have 4 or more nonzero digits in base n). This does not significantly improve the computation of a(n), however. - _Ely Golden_, Dec 30 2016
%C The pattern in the magnitude of a(n) is unclear. For some values of n, a(n) is much larger than for other values. For example, a(65) is 2460678262, whereas a(64) is only 4369 and a(66) is 4577. It seems as though even values of n typically have smaller values of a(n). - _Ely Golden_, Dec 30 2016
%C It is known that a(n) > 0 for any nonzero member of this sequence, as well as any n >= 2 of the form A280270(m), A070689(m), A279480(m), 2*A089001(m), 2*A115104(m), and 2*A280273(m)-1. It is likely, but not known, that a(n) > 0 for all n >= 2. - _Ely Golden_, Dec 30 2016
%H Ely Golden, <a href="/A278981/b278981.txt">Table of n, a(n) for n = 2..72</a> (terms a(67), a(69), and a(71) computed by Chai Wah Wu)
%H Ely Golden, <a href="/A278981/a278981_18.txt">Table of n, a(n) for n = 2..11584</a> (a-file, contains every value of a(n) <= 2^27)
%H Ely Golden, <a href="/A278981/a278981_11.txt">Proofs regarding the lower bound of A278981(n)</a>
%e a(2) = 15, as 15 is the first composite number whose base-2 nonzero digits (1111) are the same as the base-2 nonzero digits of its prime factors (11_2 and 101_2).
%t g[n_] := g[n] = Flatten[ Table[#[[1]], {#[[2]]}] & /@ FactorInteger[n]];
%t f[b_] := Block[{c = b^2}, While[ PrimeQ@ c || DeleteCases[ Sort[ IntegerDigits[c, b]], 0] != DeleteCases[ Sort[ Flatten[ IntegerDigits[g[c], b]]], 0], c++]; c]; Array[f, 39, 2] (* _Robert G. Wilson v_, Dec 30 2016 *)
%o (SageMath)
%o def nonZeroDigits(x,n):
%o if(x<=0|n<2):
%o return []
%o li=[]
%o while(x>0):
%o d=divmod(x,n)
%o if(d[1]!=0):
%o li.append(d[1])
%o x=d[0]
%o li.sort()
%o return li;
%o def nonZeroFactorDigits(x,n):
%o if(x<=0|n<2):
%o return []
%o li=[]
%o f=list(factor(x))
%o #ensures inequality of nonZeroFactorDigits(x,n) and nonZeroDigits(x,n) if x is prime
%o if((len(f)==1)&(f[0][1]==1)):
%o return [];
%o for c in range(len(f)):
%o for d in range(f[c][1]):
%o ld=nonZeroDigits(f[c][0],n)
%o li+=ld
%o li.sort()
%o return li;
%o #the actual function
%o def a(n):
%o c=2
%o while(nonZeroFactorDigits(c,n)!=nonZeroDigits(c,n)):
%o c+=1;
%o return c;
%o index=2
%o while(index<=100):
%o print(str(index)+" "+str(a(index)))
%o index+=1
%o print("complete")
%o #the actual function (alternate)
%o def a(n):
%o c=2
%o while(nonZeroFactorDigits(c,n)!=nonZeroDigits(c,n)):
%o c+=1;
%o if(c%1000000==1):
%o print("checked up to "+str(c-1))
%o return c;
%o x=3 # <some base you want to check>
%o print(str(x)+" "+str(a(x)))
%o print("complete")
%Y a(10) = A176670(1); a(2) = A278909(1).
%K nonn,base
%O 2,1
%A _Ely Golden_, Dec 02 2016
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