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A210722
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Number of ways to write n = (2-(n mod 2))p+q+2^k with p, q-1, q+1 all prime, and p-1, p+1, q all practical.
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2
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0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 3, 1, 4, 2, 4, 2, 4, 3, 4, 4, 3, 4, 5, 3, 8, 5, 8, 4, 5, 4, 5, 3, 5, 4, 4, 3, 9, 2, 12, 4, 9, 5, 7, 6, 7, 5, 5, 7, 10, 5, 13, 6, 10, 6, 8, 6, 7, 5, 1, 7, 7, 1, 10, 5, 8, 4, 9, 7, 8, 6, 3, 10, 6, 6, 10, 7, 7, 9, 11, 7, 10, 10, 5, 10, 7, 5, 10, 7, 4, 8, 8, 5, 11, 5, 8, 10, 7, 5, 12, 5
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OFFSET
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1,11
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COMMENTS
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Conjecture: a(n)>0 except for n = 1,...,8, 10, 520, 689, 740.
Zhi-Wei Sun also guessed that any integer n>6 different from 407 can be written as p+q+F_k, where p is a prime with p-1 and p+1 practical, q is a practical number with q-1 and q+1 prime, and F_k (k>=0) is a Fibonacci number.
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LINKS
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EXAMPLE
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a(1832)=1 since 1832=2*881+6+2^6 with 5, 7, 881 all prime and 6, 880, 882 all practical.
a(11969)=1 since 11969=127+11778+2^6 with 127, 11777, 11779 all prime and 126, 128, 11778 all practical.
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MATHEMATICA
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f[n_]:=f[n]=FactorInteger[n]
Pow[n_, i_]:=Pow[n, i]=Part[Part[f[n], i], 1]^(Part[Part[f[n], i], 2])
Con[n_]:=Con[n]=Sum[If[Part[Part[f[n], s+1], 1]<=DivisorSigma[1, Product[Pow[n, i], {i, 1, s}]]+1, 0, 1], {s, 1, Length[f[n]]-1}]
pr[n_]:=pr[n]=n>0&&(n<3||Mod[n, 2]+Con[n]==0)
pp[k_]:=pp[k]=pr[Prime[k]-1]==True&&pr[Prime[k]+1]==True
pq[n_]:=pq[n]=PrimeQ[n-1]==True&&PrimeQ[n+1]==True&&pr[n]==True
a[n_]:=a[n]=Sum[If[pp[j]==True&&pq[n-2^k-(2-Mod[n, 2])Prime[j]]==True, 1, 0], {k, 0, Log[2, n]}, {j, 1, PrimePi[(n-2^k)/(2-Mod[n, 2])]}]
Do[Print[n, " ", a[n]], {n, 1, 100}]
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CROSSREFS
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Cf. A005153, A208243, A208244, A208246, A208249, A209236, A209253, A209254, A209312, A209315, A209320, A210528, A210531, A210533, A210681.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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