A375256
Number of pairs of antipodal vertices in the level n Hanoi graph.
Original entry on oeis.org
3, 12, 39, 129, 453, 1677, 6429, 25149, 99453, 395517, 1577469, 6300669, 25184253, 100700157, 402726909, 1610760189, 6442745853, 25770393597, 103080394749, 412319219709, 1649272160253, 6597079203837, 26388297940989, 105553154015229, 422212540563453, 1688850011258877, 6755399743045629
Offset: 1
2 example graphs:
o
/ \
o---o
/ \
o o o
/ \ / \ / \
o---o o---o---o---o
Graph: H_1 H_2
Since the level 1 Hanoi graph is a triangle, a(1) = 3.
- Paolo Xausa, Table of n, a(n) for n = 1..1000
- Allan Bickle, Properties of Sierpinski Triangle Graphs, Springer PROMS 448 (2021) 295-303.
- A. Hinz, S. Klavzar, and S. Zemljic, A survey and classification of Sierpinski-type graphs, Discrete Applied Mathematics 217 3 (2017), 565-600.
- Eric Weisstein's World of Mathematics, Hanoi Graph
- Index entries for linear recurrences with constant coefficients, signature (7,-14,8).
Cf.
A370933 (antipodal pairs in SierpiĆski triangle graphs).
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A375256[n_] := 3*(2^(2*n - 3) + 3*2^(n - 2) - 1);
Array[A375256, 30] (* or *)
LinearRecurrence[{7, -14, 8}, {3, 12, 39}, 30] (* Paolo Xausa, Sep 23 2024 *)
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a(n) = 3*(2^(2*n-3)+3*2^(n-2)-1); \\ Michel Marcus, Aug 08 2024
A296807
Take a prime, convert it to base 2. Consider it as a string of digits and delete its leftmost and rightmost digit. Leading zeros are kept. Repeat the process. a(n) is the least prime that, in the first n steps of this process, generates a string that is a prime read in base 2.
Original entry on oeis.org
2, 13, 43, 151, 2143, 2143, 12479, 57727, 246527, 4267455487, 276009615632383, 4469780781584383, 576406542684520447
Offset: 0
a(1) = 13 because 13 in base 2 is 1101 and 10 is 2 and 13 is the least number with this property;
a(2) = 43 because 43 in base 2 is 101011 while 0101 is 5 and 10 is 2 and 43 is the least number with this property;
a(3) = 151 because 151 in base 2 is 10010111 while 001011 is 11, 0101 is 5 and 10 is 2 and 151 is the least number with this property.
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with(numtheory): P:=proc(q) local a,b,c,i,j,k,n,ok,x; x:=5; for k from 1 to q do for n from x to q do a:=convert(ithprime(n),base,2); ok:=1; for i from 1 to k do b:=nops(a)-i; while a[b]=0 do b:=b-1; od;
c:=0; for j from b by -1 to i+1 do c:=2*c+a[j]; od;if not isprime(c) then ok:=0; break; fi; od;if ok=1 then x:=n; print(ithprime(n)); break; fi; od; od; end: P(10^20);
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Table[SelectFirst[Prime@ Range[#, # + 10^5] &@ PrimePi[2 (4^n + 2^n) + 1], AllTrue[Map[FromDigits[#, 2] &, Rest@ NestWhileList[Most@ Rest@ # &, IntegerDigits[#, 2], Length@ # > 2 &]], PrimeQ] &], {n, 8}] (* Michael De Vlieger, Dec 29 2017 *)
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a(n) = if(!n, return(2)); forprime(p=2*4^n + 3*2^n - 1, , my(b=p); for(x=1, n, b = (b - (b>=4*2^(logint(p, 2) - 2*x))*4*2^(logint(p, 2) - 2*x) - 1)/2; if(!isprime(b) || (b==2 && x!=n), next(2))); return(p)) \\ Iain Fox, Dec 29 2017 (corrected by Iain Fox, Oct 26 2019)
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