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.
f:= proc(n) local k,p;
p:= n;
for k from 1 do
p:= p*(n+k);
if (p/(n+k+1))::integer then return k fi
od
end proc:
map(f, [$1..100]); # Robert Israel, Feb 25 2020
a[n_] := Module[{k, p = n}, For[k = 1, True, k++, p *= (n+k); If[Divisible[p, n+k+1], Return[k]]]];
Array[a, 100] (* Jean-François Alcover, Jun 04 2020, after Maple *)
a(n) = {my(r=n*(n+1)); for(k=2, oo, r=r*(n+k); if(r%(n+k+1)==0, return(k))); } \\ Jinyuan Wang, Feb 25 2020
\\ See Corneth link
def a(n):
k, p = 1, n*(n+1)
while p%(n+k+1): k += 1; p *= (n+k)
return k
print([a(n) for n in range(1, 85)]) # Michael S. Branicky, Jun 06 2021
k = 13 is not a term since 12345678910111213 is not divisible by 13.
b = 10; c = {}; Select[Range[10^5], Divisible[FromDigits[c = Join[c, IntegerDigits[#, b]], b], #] &] (* Robert Price, Mar 11 2020 *)
Select[Range[1200],Divisible[FromDigits[Flatten[IntegerDigits/@Range[#]]],#]&] (* Harvey P. Dale, Dec 31 2020 *)
nxt[{rc_,n_}]:={rc*10^IntegerLength[n+1]+n+1,n+1}; Select[NestList[nxt,{1,1},1200],Mod[#[[1]],#[[2]]]==0&][[;;,2]] (* Harvey P. Dale, Sep 26 2023 *)
c=0;for(d=1,1e9,for(n=d,-1+d*=10,(c=c*d+n)%n || print1(n","));d--) \\ M. F. Hasler, Sep 11 2011
A029455_list, r = [], 0 for n in range(1,10**4+1): r = r*10**len(str(n))+n if not (r % n): A029455_list.append(n) # Chai Wah Wu, Nov 05 2014
def concat_mod(base, k, mod):
total, digits, n1 = 0, 1, 1
while n1 <= k:
n2, p = min(n1*base-1, k), n1*base
# Compute ((p-1)*n1+1)*p**(n2-n1+1)-(n2+1)*p+n2 divided by (p-1)**2.
# Since (a//b)%mod == (a%(b*mod))//b, compute the numerator mod (p-1)**2*mod.
tmp = pow(p, n2-n1+1, (p-1)**2*mod)
tmp = ((p-1)*n1+1)*tmp-(n2+1)*p+n2
tmp = (tmp%((p-1)**2*mod))//(p-1)**2
total = (total*pow(p, n2-n1+1, mod)+tmp)%mod
digits, n1 = digits+1, p
return total
for k in range(1, 10**10+1):
if concat_mod(10, k, k) == 0: print(k) # Jason Yuen, Jan 27 2024
a(2) = 2 as 2*2 = 4 which contains 2 + 2 = 4 as a substring. a(4) = 68 as 4*68 = 272 which contains 4+68 = 72 as a substring. a(69) = 16961 as 69*16961 = 1170309 which contains 69+16961 = 17030 as a substring. a(501000) = 1002 as 501000*1002 = 502002000 which contains 501000+1002 = 502002 as a substring. This is the first of 500 consecutive terms with a(n) = 1002. a(554635) = 879948670 as 554635*879948670 = 488050330585450 which contain 554635+879948670 = 880503305 as a substring. This is the largest value of a(n) for the first one million terms.
isok(n, k) = #strsplit(Str(n*k), Str(n+k)) > 1;
a(n) = {if (vecsearch([1, 3, 5, 6, 7, 9, 26], n), return (-1)); my(k=1); while (! isok(k, n), k++); k;} \\ Michel Marcus, Dec 02 2020 and Jan 23 2021
a(1) = 2 as '1' || '2' = '12', which is divisible by 3 (where || denotes decimal concatenation). a(7) = 20 as '7' || '8' || '9' || '10' || '11' || '12' || ... || '20' = 7891011121314151617181920, which is divisible by 21. a(8) = 10 as '8' || '9' || '10' = 8910, which is divisible by 11. a(2) = 82: the concatenation 2 || 3 || ... || 82 is 23456789101112131415161718192021222324252627282930313233343536373839\ 40414243444546474849505152535455565758596061626364656667686970717273747\ 576777879808182, which is divisible by 83.
grow := proc(n,M) # searches out to a limit of M, returns [n,n+k] or [n,-1] if no k was found local R,i; R:=n; for i from n+1 to M do R:=R*10^length(i)+i; if (i mod 2) = 0 then if (R mod (i+1)) = 0 then return([n, i]); fi; fi; od: [n, -1]; end; for n from 1 to 100 do lprint(grow(n,20000)); od;
apply( {A332584(n,L=10^#Str(n),c=n)= until((c=c*L+n)%(n+1)==0, n++M. F. Hasler, Feb 20 2020
def A332584(n): r, m = n, n + 1 while True: r = r*10**(len(str(m))) + m if m % 2 == 0 and r % (m+1) == 0: return m m += 1 # Chai Wah Wu, Jun 12 2020
3 divides 12, 9 divides 12345678.
s = ""; Do[s = s <> ToString[n]; If[Mod[ToExpression[s], n + 1] == 0, Print[n + 1]], {n, 0, 5*10^6}] (* Ryan Propper, Aug 28 2005 *)
Select[Range[55*10^5],Divisible[FromDigits[Flatten[IntegerDigits/@Range[ #-1]]],#]&] (* Harvey P. Dale, Mar 28 2020 *)
# See A029455 for concat_mod def isok(k): return concat_mod(10, k-1, k)==0 # Jason Yuen, Oct 06 2024
a(3) = 24 as 3*24 = 72 which contains reverse(3+24) = reverse(27) = 72 as a substring. a(6) = 34 as 6*34 = 204 which contains reverse(6+34) = reverse(40) = 04 as a substring. Note the leading zero is included. a(29) = 46716 as 29*46716 = 1354764 which contains reverse(29+4671) = reverse(46745) = 54764 as a substring. a(110) = 11 as 110*11 = 1210 which contains reverse(110+11) = reverse(121) = 121 as a substring. This is the first of two consecutive terms with a(n) = 11. a(20000) = 666843331 as 20000*666843331 = 13336866620000 which contains reverse(20000+666843331) = reverse(666863331) = 133368666 as a substring. This is the largest value in the first 100000 terms.
isok(n, k) = #strsplit(Str(n*k), concat(Vecrev(Str(n+k)))) > 1;
ispt(n) = my(t); ispower(n,,&t) && (t==10);
a(n) = {if ((n==1) || (n==10) || ispt(n), return (-1)); my(k=0); while (! isok(n, k), k++); k;} \\ Michel Marcus, Jan 22 2021
a(1) = 1 as prime(1) + prime(2) = 2 + 3 = 5, which is divisible by prime(3) = 5. a(4) = 11 as prime(4) + ... + prime(15) = 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 43 + 47 = 318, which is divisible by prime(16) = 53.
from sympy import prime, nextprime def A360297(n): p = prime(n) q = nextprime(p) s, k = p+q, 1 while s%(q:=nextprime(q)): k += 1 s += q return k # Chai Wah Wu, Feb 03 2023
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